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Net Ionic Equation for HNO3 + NaOH (Strong Base and Strong Acid)
The reaction of Nitric acid and Sodium hydroxide represents a net ionic equation involving a strong acid and strong base. Strong acids and strong bases are considered strong electrolytes and will dissociate completely. This means that we will split them apart in the net ionic equation.
HNO3 + NaOH = NaNO3 + H2O is a neutralization reaction (also a double displacement reaction). The strong acid (HNO3) and strong base react to produce a salt (NaNO3) and water (H2O).
Video: HNO3 + NaOH (Net Ionic Equation)
To balance net ionic equations we follow these general rules:
Write the balanced molecular equation.
Write the state (s, l, g, aq) for each substance.
Split strong electrolytes into ions (the complete ionic equation).
Cross out the spectator ions on both sides of complete ionic equation.
Write the remaining substances as the net ionic equation.
Note: charges in a net ionic equation are conserved. This means that the overall charge (called the net charge) on the reactants side (left) of the equation must equal to the net charge on the products side (right).
Video: How to Balance Net Ionic Equations
More Worked examples of Net Ionic Equations
Net Ionic Equation Writing (from Formulas) – (Spectator ions appear the same on both sides of the reaction.) (They don't actually participate in the reaction.) (If you did this correctly the ions Write balanced formula unit, total ionic and net ionic equations for the following reactions. Assume all reactions occur in water or in contact with water.Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. The chemical equation for the reaction of nitric acid and sodium hydroxide is given asIn this video we determine the type of chemical reaction for the equation HNO3 + NaOH = NaNO3 + H2O (Nitric acid + Sodium hydroxide).Since we have an acid…
Which equation represents the total ionic equation for the reaction… – Write the total ionic equation for the reaction of hydrofluoric acid with potassium hydroxide. Give the net ionic equation for the reaction (if any) that occurs when aqueous solutions of h2so4 a…Solved and balanced chemical equation NaOH + HNO3 → NaNO3 + H2O with completed products. Application for completing products and balancing equations. This is an acid-base reaction ( neutralization ): NaOH is a base , HNO3 is an acid .Now, the complete ionic equation looks like this — keep in mind that sodium fluoride is soluble in aqueous solution. Notice that because hydrofluoric acid does not dissociate completely, you must represent it in molecular form. Eliminate the spectator ions, which in this case are the sodium cations.
Type of Reaction for HNO3 + NaOH = NaNO3 + H2O – YouTube – A Net Ionic Equation is a chemical equation for a reaction which lists only those species participating in the reaction.To write a Net Ionic Reaction, follow these 3 steps:1) Start by simply writing the overall balanced chemical reaction. This is also called the Molecular Equation.2) Then…The equation he gives represents the stoichiometry used in my buffer calculations, with the Dear Sir. Concerning your question about the reaction of boric acid with sodium hydroxide. The speciation in this system is complex, di- and triborate ions are present in addition to the monoborate ion.Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below. Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co – cobalt and CO – carbon monoxide.
3A 14.4 – .
Hess's Law – Chemistry Tutorial – hello and welcome to the chemistry
solution this tutorial is on Hess's law remember that enthalpy is a state
function and that means it depends only on the current state of the system that
means that the enthalpy change are Delta H for a reaction depends only on the
initial and final conditions and not the path taken and so whether a given
reaction occurs in one step or in many steps the enthalpy change remains the
same and this is the basis for Hess's law
Hess's law states that the enthalpy change associated with a reaction
carried out in multiple steps is equal to the sum of the enthalpy changes for
each individual step so let's look at an example calculate Delta H for the
following reaction to carbon monoxide plus o2 goes to two carbon dioxide and
we'll calculate Delta H for this reaction given the two reactions below
and their Delta H values so the first step to do this is to manipulate the
given equations so that they most closely resemble the equation of
interest so let's look at our first equation in the reaction of interest
carbon monoxide is on the reactant side and of the two equations given below you
only see carbon monoxide in one of these equations and it's on the product side
so what we're going to need to do is to flip this equation around so the carbon
monoxide is on the reactant side so we'll write the equation below except
we'll flip it around so here we have two carbon monoxide going to two carbon plus
o2 and because we flipped this reaction we need to flip the sign on Delta H so
remember the Delta H value for a reaction written in the opposite
direction is of the same magnitude but of opposite sign and so if we flip the
equation we flip the sign on Delta H of to carbon plus o2 gives us two co and
the Delta H value is negative 221 kilojoules per mole then 2 Co going to 2
carbon plus o2 has a delta H value of positive 221 kilojoules per mole
looking at our reaction of interest I see that there are two moles of co2 on
the product side and when I look down at our given equations I see only one of
the equations have co2 the second one it is on the product side but there's only
one mole of co2 in this reaction and so for this equation I need to multiply it
by 2 so that I have 2 moles of co2 so it more closely resembles the reaction of
interest and when I multiply the reaction by 2 I need to multiply the
whole equation by 2 and I think it helps to draw a parentheses around the entire
equation so this gives us 2 carbon plus 2 O 2 gives us 2 co2
and again I want to perform the same operation on Delta H as I did on the
equation so if I multiplied the equation by 2 I need to multiply the Delta H
value by 2 so if the reaction carbon plus O 2 gives us co2 had a Delta H
value of negative 300 93.5 kilojoules per mole then the reaction to carbon
plus 2 O 2 gives us 2 co 2 has a Delta H value of negative 787 kilojoules per
mole so here I've written my two new equations that I modified to more
closely resemble my equation of interest and what you're going to do now is add
these two equations together keeping everything on the reactant side on the
reactant side and everything on the product side on the product side so in
my new equation I have to Co plus 2 C plus 2 O 2 gives me 2 c plus o 2 plus 2
co 2 and because I added the reactions
together I'm going to add the Delta H values together so the Delta H value for
this equation is negative 566 kilojoules per mole but you'll notice that I have
some compounds that are the same on both the reactant side and the product side
and so anything that's the same on both sides of the reaction arrow can be
canceled out so I can cancel out my two moles of carbon on both the reactants
and the products side and I can also cancel out the one mole of ol2 on the
product side with one mole of o2 on the reactant side leaving me with the final
equation to Co plus o2 goes to two co2 and the last thing you need to do is to
double check that this equation matches your equation of interest and in this
case it does so the Delta H value for two CL + o2 goes to 2 CO 2 is negative
566 kilojoules per mole let's try another example calculate
Delta H for the following reaction – n2 + 5 o2 goes to 2 and 205 we're given
three reactions listed below and remember the first step is to modify
these equations to more closely resemble the equation of interest now I think
it's always helpful to look for a compound in the reaction of interest
that is found only once in your given equations and in this case you can see
that n 2 O 5 that's on the product side and our reaction of interest isn't only
one of the given equations and you'll see that it's on the reactant side so
the first thing we need to do is to flip this equation around so that n 205 is on
the product side the other thing we need to do is to multiply this equation by 2
because our reaction of interest has two moles of n2o5 on the product side and
remember we perform the same operation on Delta H as we did on the equate
Asian so if we flip the equation and multiply it by two we need to flip the
sign on Delta H and multiply it by two this gives us the new reaction for hno3
goes to two n 2o 5 + 2 h2o and this reaction has a delta h value of
153 kilojoules per mole looking at the bottom equation I'll see
that I have n 2 on the reactant side and in my reaction of interest I also have n
2 on your Acton side but in my reaction of interest I have 2 moles of n2 on the
reactant side so in order to make this equation more closely resemble a
reaction of interest I need to multiply the entire equation by 2 now you'll also
notice in this equation that two moles of h no.3 are formed on the product side
and we compare that to the equation that we just modified we have 4 moles of h
no.3 on the reactant side if you look at our reaction of interest up on the top
you don't see the compound hno3 anywhere in that equation which means that you're
going to need equal amounts of hno3 on the reactant side and the product side
in order to cancel out in our final equation so we already knew that we
needed to multiply this equation by 2 to give us 2 moles of n2 but you can also
look at it that we need to multiply this equation by 2 to give us 4 moles of h
no.3 which will eventually cancel out with the 4 moles of h no.3 on the
reactant side in our first equation that we modified and when we multiply the
equation by 2 we also need to multiply the Delta H value by 2 now I save this
top equation for last and it's because neither h2 nor h2o are
anywhere to be found in our equation of interest and so we need to compare this
reaction to the reactions that we already modified in order to figure out
how to modify this equation in order to cancel out any moles of h2o or h2 gas in
our final equation and what you'll see is that
we have two moles of h2o on the product side and we need to modify this third
equation in order to cancel those out so we'll need to flip this equation and
multiply it by two in order to cancel out h2o by having it on both the
reactants and the products side and you see the same thing with h2 gas so when
we flip this equation and multiply it by two remember that we need to do the same
thing for Delta H flip the sign on Delta H and multiply it by two now that we
have two moles of water on the reactant side and the product side those will
cancel out when we add these equations together and the same thing with our two
moles of h2 on the reactant side and the product side now hopefully we've
modified all of our equations correctly so remember we add everything on the
reactant side and everything on the product side and then we can cancel out
anything that's the same on both sides now if you don't want to write all of
this out you can cancel things that are the same
on the reactant side and the product side right away before adding your
equations together and so you'll see that we have four moles of H no.3 on the
reactant side and the product side we have two moles of h2o on the reactant
side and the product side and we have two moles of h2 on the reactant side in
the product side our one mole of o2 gas on the product side can cancel with one
mole of o2 gas on the reactant side leaving us with five moles of o2 on the
reactant side and when we write down what we're left with we then have an
additional two moles of n2 on the reactant side and two moles of n2o5 on
the product side and when you check our new equation now matches the equation of
interest and so because we added up these three new equations we also need
to add up their Delta H values and so the Delta H value for this reaction 2 +
2 + 502 goes to 2n 205 is 29 kilojoules per mole thanks for watching the chemistry
solution we hope you enjoyed this tutorial you .
Neutralization (chemistry) – In chemistry, neutralization or neutralisation,
is a chemical reaction in which an acid and a base react quantitatively with each other.
In a reaction in water neutralization results in there being no excess of hydrogen or hydroxide
ions present in solution.
The pH of the neutralized solution depends on the acid strength of the
reactants. Neutralization is used in many applications. Meaning of "neutralisation"
In the context of a chemical reaction the term neutralisation is used for a reaction
between an acid and a base or alkali. Historically, this reaction was represented as
acid + base = salt + water. For example, HCl + NaOH → NaCl + H2O
The statement is still valid as long as it is understood that in aqueous solution the
substances involved are subject to dissociation. The arrow sign, →, is used because the reaction
is complete, that is, neutralisation is a quantitative reaction. A more general definition
is based on Brønsted–Lowry acid–base theory.
AH + Bz → A- +BHz+1 AH represents an acid and B represents a base.
z is an electric charge; negative for an anion, zero, or positive for a cation. When the reaction
takes place in water and the base is the hydroxide ion, OH-, the reaction can be written as
AH + OH- → A- + H2O, When the acid has been neutralised there are
no molecules of AH left in the solution. It follows also that, in a neutralization reaction,
the amount of base added must be equal to the amount of acid present initially. This
amount of base is said to be the equivalent amount. In a titration of an acid with a base,
the point of neutralization can also be called the equivalence point. The quantitative nature
of the neutralisation reaction is most conveniently expressed in terms of the concentrations of
acid and alkali. At the equivalence point volume × concentration = volume × concentration
In general, for an acid AHn at concentration c1 reacting with a base B(OH)m at concentration
c2 the volumes are related by n v1 c1 = m v2 c2
The concept of neutralisation is not limited to reactions in solution. For example, the
reaction of limestone with acid such as sulfuric acid is also
a neutralization reaction.
[Ca,Mg]CO3(s) + H2SO4(aq) → [Ca,Mg]SO4(s) + CO2(g) + H2O
Such reactions are important in soil chemistry. Strong acids and strong bases
A strong acid is one that is fully dissociated in aqueous solution. For example hydrochloric
acid, HCl, is a strong acid. HCl(aq) → H+(aq) + Cl-(aq)
A strong base is one that is fully dissociated in aqueous solution. For example sodium hydroxide,
NaOH, is a strong base. NaOH(aq) → Na+(aq) + OH-(aq)
Therefore when a strong acid reacts with a strong base the neutralization reaction can
be written as H+ + OH- → H2O
For example, in the reaction between hydrochloric acid and sodium hydroxide the sodium and chloride
ions, Na+ and Cl- take no part in the reaction. The reaction is consistent with the Brønsted–Lowry
definition because in reality the hydrogen ion exists as the hydronium ion, so that the
neutralization reaction may be written as H3O+ + OH- → H2O + H2O
When a strong acid is neutralized by a strong base there are no excess hydrogen ions left
in the solution. The solution is said to be neutral as it is neither acidic nor alkaline.
The pH of such a solution is close to a value of 7; the exact pH value is dependent on the
temperature of the solution. Neutralization is an exothermic reaction.
The standard enthalpy change for the reaction H+ + OH- → H2O is -55.90 kJ/mol.
Weak acids and strong bases A weak acid is one that does not dissociate
fully when it is dissolved in water. Instead an equilibrium mixture is formed.
AH + H2O H3O+ + A- Acetic acid is an example of a weak acid.
The pH of the neutralized solution is not close to 7, as with a strong acid, but depends
on the acid dissociation constantof the acid. The pH at the end-point or equivalence point
in a titration may be easily calculated. At the end-point the acid is completely neutralized
so the analytical hydrogen ion concentration, TH, is zero and the concentration of the conjugate
base, A-, is effectively equal to the analytical concentration of the acid; writing AH for
the acid, [A-]= TA. Defining the acid dissociation constant, pKa, as [HA] = Ka[A-][H+]; pKa = -log10Ka
and the self-dissociation constant for water, Kw, as
Kw = [H+][OH-]; pKw = -log10Kw the equation for mass-balance in hydrogen
ions is easy to write down. TH= [H+] +Ka[A-][H+] – Kw[H]-1
The term Kw[H]−1 is equal to the concentration of hydroxide ions. At neutralization, TH is
zero. [H+] + Ka[A][H+] – Kw[H ]-1 = 0
[H+]2 + KaTA[H+]2 – Kw = 0 [H+]2 = Kw /
log [H+] = 1/2 log Kw – 1/2 log(1 + KaTA) pH= 1/2 pKw – 1/2 log(1 + TA/Ka)
In most circumstances the term 1+TA/Ka is much larger than 1 and is equal to TA/Ka to
a good approximation. pH ≈ 1/2
This equation explains explains the following facts:
The pH at the end-point depends mainly on the strength of the acid, pKa.
The pH at the end-point also depends on the concentration of the acid, TA.
The pH rises more steeply at the end-point as the acid concentration increases.
When a weak acid is titrated with a strong base the end-point occurs at pH greater than
7. Therefore, the most suitable indicator to use is one, like Phenolphthalein, that
changes color at high pH. Weak bases and strong acids
The situation is analogous to that of weak acid and strong base.
H3O+ + B H2O + BH+ The pH of the neutralized solution depends
on the acid dissociation constant of the base, pKa, or, equivalently, on the base association
constant, pKb. The most suitable indicator to use for this
type of titration is one, like Methyl orange, that changes color at low pH.
Weak acids and weak bases When a weak acid reacts with an equivalent
amount of a weak base complete neutralization does not occur.
AH + B A- + BH+ The concentrations of the species in equilibrium
with each other will depend on the equilibrium constant, K, for the reaction, which can be
defined as follows. [A-][BH+] = K[AH][B]
Given the association constants for the acid and the base.
A- + H+ AH; [AH] = Ka[A-][H+] B + H+ BH+; [BH+] = Kb[B][H+]
it follows that K = Ka / Kb A weak acid cannot be neutralized by a weak
base, and vice versa.. Applications
Chemical titration methods are used for analyzing acids or bases to determine the unknown concentration.
Either a pH meter or a pH indicator which shows the point of neutralization by a distinct
color change can be employed. Simple stoichiometric calculations with the known volume of the
unknown and the known volume and molarity of the added chemical gives the molarity of
the unknown. In wastewater treatment, chemical neutralization
methods are often applied to reduce the damage that an effluent may cause upon release to
the environment. For pH control, popular chemicals include calcium carbonate, calcium oxide,
magnesium hydroxide, and sodium bicarbonate. The selection of an appropriate neutralization
chemical depends on the particular application. There are many uses of neutralization reactions
that are acid-alkali reactions. A very common use is antacid tablets. These are designed
to neutralize excess gastric acid in the stomach that may be causing discomfort in the stomach
or lower esophagus. This can also be remedied by the ingestion of sodium bicarbonate.
Also in the digestive tract, neutralization reactions are used when food is moved from
the stomach to the intestines. In order for the nutrients to be absorbed through the intestinal
wall, an alkaline environment is needed, so the pancreas produce an antacid bicarbonate
to cause this transformation to occur. Another common use, though perhaps not as
widely known, is in fertilizers and control of soil pH. Slaked lime or limestone may be
worked into soil that is too acidic for plant growth. Fertilizers that improve plant growth
are made by neutralizing sulfuric acid or nitric acid with ammonia gas, making ammonium
sulfate or ammonium nitrate. These are salts utilized in the fertilizer.
Industrially, a by-product of the burning of coal, sulfur dioxide gas, may combine with
water vapor in the air to eventually produce sulfuric acid, which falls as acid rain. To
prevent the sulfur dioxide from being released, a device known as a scrubber gleans the gas
from smoke stacks. This device first blows calcium carbonate into the combustion chamber
where it decomposes into calcium oxide and carbon dioxide. This lime then reacts with
the sulfur dioxide produced forming calcium sulfite. A suspension of lime is then injected
into the mixture to produce a slurry, which removes the calcium sulfite and any remaining
unreacted sulfur dioxide. Further reading
Neutralization is covered in most general chemistry text-books. Detailed treatments
may be found in text-books on analytical chemistry such as
Skoog, D.A; West, D.M.; Holler, J.F.; Crouch, S.R.. Fundamentals of Analytical Chemistry.
Thomson Brooks/Cole. ISBN 0-03-035523-0. Chapters 14, 15 and 16
Applications Stumm, W.; Morgan, J.J.. Water Chemistry.
New York: Wiley. ISBN 0-471-05196-9. Snoeyink, V.L.; Jenkins, D.. Aquatic Chemistry:
Chemical Equilibria and Rates in Natural Waters. New York: Wiley. ISBN 0-471-51185-4.
Millero, F.J.. Chemical Oceanography. London: Taylor and Francis. ISBN 0-8493-2280-4.
Metcalf & Eddy. Wastewater Engineering, Treatment and Reuse. 4th ed. New York: McGraw-Hill,
2003. 526-532. References
^ Jarvis, Alan & Beavon, Rod. "Periodicity Quantitative Equilibria and Functional Group
Chemistry", 16 January 2001. ^ Steven S. Zumdahl. Chemical Principles.
New York: Houghton Mifflin Company. pp. 319–324. .