Equal volumes of 0.594 M aqueous solutions of nitrous acid ( HNO 2 (aq) ) and so
ID: 499234 • Letter: E
Question
Equal volumes of 0.594 M aqueous solutions of nitrous acid (HNO2(aq)) and sodium benzoate (NaC6H5COO) are mixed.
(1) Write the net ionic equation for the overall reaction that takes place as the system comes to equilibrium.
Write acetic acid, benzoic acid or formic acid and their conjugates in the form RCOOH / RCOO-. For example, benzoic acid should be written "C6H5COOH" NOT "C6H5CO2H". It is not necessary to include states such as (aq).
(2) What is the value of the equilibrium constant for this reaction?
(3) What is the pH of the resulting solution?
Equilibrium constants found here: https://docs.google.com/document/d/1cNVOoEOVMqPSx8QdL8mc3zd8yFTwBx8VWd2MiDCDA1o/edit
_____ + _____ ---> _____ + _____Explanation / Answer
Answer:
1)
HNO2 <----------> H+ (aq) + NO2- (aq)
C6H5COONa ------------> C6H5COO- (aq) + Na+ (aq)
On adding these separate ionization equations,
H+ + NO2- + C6H5COO- + Na+ -----------> C6H5COOH + NaNO2.
H+ + NO2- + C6H5COO- + Na+ -----------> C6H5COOH + Na+ + NO2-. (NaNO2 being strong electrolyte)
HNO2 is relatively stronger acid than Benzoic acid so it is the Benzoic acid that will be predominant species on mixing. (see corresponding Ka values from table you provided in Q.3)
Na+ and NO2-. mutually cancelled out from both sides,
H+ + C6H5COO- -----------> C6H5COOH
Is the nett Ionic equation.
2) The Net ionic equation is,
H+ + C6H5COO- -----------> C6H5COOH
This eqution is about basicity of a conjugate base C6H5COO- . it's dissociation constant Kb is given as,
Kb = [H+][C6H5COO-] / [C6H5COOH]
Ka for benzoic acid is = 6.3 x 10-5.
Ka x Kb = Kw = 1 x 10-14.
Kb = (1 x 10-14) / (6.3 x 10-5)
Kb = 1.6 x 10-10.
Hence equilibrium constant for reaction in (1) is given as,
Kb = [H+][C6H5COO-] / [C6H5COOH] = 1.6 x 10-10. ------------- (I)
3) As there is equimolar concentrations of Acid and conjugate base componant used,
pH = pKa --------- (by Henderson Hasselbalch equation)
pH = -log(Ka)
pH = -log(1.6 x 10-10)
pH = 9.8
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pH = -log(
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