## Chemistry (7th Edition)

Published by Pearson

# Chapter 16 - Applications of Aqueous Equilibria - Section Problems - Page 710: 70

#### Answer

$pH = 9.09$ The pH of the solution will not be affected if diluted by a factor of 2, because, considering the approximation that "x" is small, if both concentrations are divided (or multiplied) by 2, the ratio will still the same, and the pH will be the same as well.

#### Work Step by Step

1. Drawing the ICE table, we get these concentrations at the equilibrium: $HCN(aq) + H_2O(l) \lt -- \gt CN^-(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[HCN] = 0.2 M - x$ $[CN^-] = 0.12M + x$ $[H_3O^+] = 0 + x$ 2. Calculate 'x' using the $K_a$ expression. $4.9\times 10^{- 10} = \frac{[CN^-][H_3O^+]}{[HCN]}$ $4.9\times 10^{- 10} = \frac{( 0.12 + x )* x}{ 0.2 - x}$ Considering 'x' has a very small value. $4.9\times 10^{- 10} = \frac{ 0.12 * x}{ 0.2}$ $4.9\times 10^{- 10} = 0.6x$ $\frac{ 4.9\times 10^{- 10}}{ 0.6} = x$ $x = 8.2\times 10^{- 10}$ Percent dissociation: $\frac{ 8.2\times 10^{- 10}}{ 0.2} \times 100\% = 4.1\times 10^{- 7}\%$ x = $[H_3O^+]$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 8.2 \times 10^{- 10})$ $pH = 9.09$ ----- $[HCN] = 0.1M$ and $[CN^-] = 0.06M$ $4.9\times 10^{- 10} = \frac{ 0.06 * x}{ 0.1}$ $4.9 \times 10^{-10} = 0.6x$ $x = 8.2 \times 10^{-10}M$ Therefore, the hydronium concentration will still the same (considering the "x" is small approximation), which makes the pH equal too.

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