Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 16 - Sections 16.1-16.8 - Exercises - Problems by Topic - Page 805: 59b

Answer

The $KOH$ did not exceed the buffer capacity.

Work Step by Step

1. Calculate the molar mass $(KOH)$: 39.1* 1 + 16* 1 + 1.01* 1 = 56.11g/mol 2. Calculate the number of moles $(KOH)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{0.35}{ 56.11}$ $n(moles) = 6.2\times 10^{- 3}$ 3. Find the concentration in mol/L $(KOH)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $ C(mol/L) = \frac{ 6.2\times 10^{- 3}}{ 0.5} $ $C(mol/L) = 0.012$ 4. Drawing the ICE table we get these concentrations at the equilibrium: $HNO_2(aq) + H_2O(l) \lt -- \gt N{O_2}^-(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[HNO_2] = 0.1 M - x$ $[N{O_2}^-] = 0.15M + x$ $[H_3O^+] = 0 + x$ 5. Calculate 'x' using the $K_a$ expression. $ 4.6\times 10^{- 4} = \frac{[N{O_2}^-][H_3O^+]}{[HNO_2]}$ $ 4.6\times 10^{- 4} = \frac{( 0.15 + x )* x}{ 0.1 - x}$ Considering 'x' has a very small value. $ 4.6\times 10^{- 4} = \frac{ 0.15 * x}{ 0.1}$ $ 4.6\times 10^{- 4} = 1.5x$ $\frac{ 4.6\times 10^{- 4}}{ 1.5} = x$ $x = 3.1\times 10^{- 4}$ Percent dissociation: $\frac{ 3.1\times 10^{- 4}}{ 0.1} \times 100\% = 0.31\%$ x = $[H_3O^+]$ $[HNO_2] = 0.1 M - x = 0.1 M - 3.1 \times 10^{-4}M \approx 0.1M$ $[N{O_2}^-] = 0.15M + x = 0.1 M + 3.1 \times 10^{-4}M \approx 0.15M$ $[H_3O^+] = 0 + x = 3.1 \times 10^{-4}M$ 6. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 3.1 \times 10^{- 4})$ $pH = 3.51$ 7. Since we are adding a strong base, this reaction will occur: $HNO_2(aq) + OH^-(aq) -- \gt N{O_2}^-(aq) + H_2O(l)$ And these are the concentrations after this reaction: Remember: Reactants at equilibrium = Initial Concentration - y And Products = Initial Concentration + y Since $KOH$ is a strong base, y = $[KOH] = 0.012M$ $[HNO_2] = 0.1 M - 0.012 = 0.088M$ $[N{O_2}^-] = 0.15M + 0.012 = 0.16M$ 8. Now, calculate the hydronium ion concentration after the addition of the $KOH$: $[H_3O^+] = Ka * (\frac{[HNO_2]}{[N{O_2}^-]})$ $[H_3O^+] = 4.7 \times 10^{-4} * \frac{0.088}{0.16}$ $[H_3O^+] = 4.7 \times 10^{-4} * 0.54$ $[H_3O^+] = 2.5 \times 10^{-4}$ 9. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 2.5 \times 10^{- 4})$ $pH = 3.60$ 10. Calculate the pH difference: $3.60 - 3.51 = 0.09$ Since the difference is smaller than 1, we consider that, this compound did not exceed the buffer capacity.
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