Chemistry and Chemical Reactivity (9th Edition)

Hydronium ion concentration: $[H_3O^+] = 1.1 \times 10^{- 8}M$ $pH = 7.96$
Analyze each ion that forms that salt: $Na^+$: Conjugate acid of $NaOH$ (strong base). Therefore, it will not affect the pH of the water. $HCO_2^-$: Conjugate base of a weak acid $(HCO_2H)$. Calculate the pH for a $0.015M$ $HCO_2^-$ solution. - Since $HCO_2^-$ is the conjugate base of $HCO_2H$ , we can calculate its $K_b$ using this equation: $K_a * K_b = K_w = 10^{-14}$ $1.8\times 10^{- 4} * K_b = 10^{-14}$ $K_b = \frac{10^{-14}}{ 1.8\times 10^{- 4}}$ $K_b = 5.6\times 10^{- 11}$ 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $HCO_2^-(aq) + H_2O(l) \lt -- \gt HCO_2H(aq) + OH^-(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $HCO_2H$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M -$[OH^-] = [HCO_2H] = 0 + x = x$ -$[HCO_2^-] = [HCO_2^-]_{initial} - x$ For approximation, we are going to consider $[HCO_2^-]_{initial} = [HCO_2^-]$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][HCO_2H]}{ [HCO_2^-]}$ $Kb = 5.6 \times 10^{- 11}= \frac{x * x}{ 0.015}$ $Kb = 5.6 \times 10^{- 11}= \frac{x^2}{ 0.015}$ $x^2 = 5.6 \times 10^{-11} \times 0.015$ $x = \sqrt { 5.6 \times 10^{-11} \times 0.015} = 9.2 \times 10^{-7}$ Percent ionization: $\frac{ 9.2 \times 10^{- 7}}{ 0.015} \times 100\% = 0.0061\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [HCO_2H] = x = 9.2 \times 10^{- 7}M$ $[HCO_2^-] \approx 0.015M$ 3. Calculate the hydronium ion concentration: $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $9.2 \times 10^{- 7} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 9.2 \times 10^{- 7}}$ $[H_3O^+] = 1.1 \times 10^{- 8}M$ 5. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.1 \times 10^{- 8})$ $pH = 7.96$