## Chemistry: The Molecular Science (5th Edition)

$pH = 7.23$
1. Calculate the molar mass $(NaOH)$: 22.99* 1 + 16* 1 + 1.01* 1 = 40g/mol 2. Calculate the number of moles $(NaOH)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.425}{ 40}$ $n(moles) = 0.0106$ 3. Find the concentration in mol/L $(NaOH)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.0106}{ 2}$ $C(mol/L) = 5.31\times 10^{- 3}$ 4. Drawing the ICE table, we get these concentrations at the equilibrium: $H_2P{O_4}^-(aq) + H_2O(l) \lt -- \gt HP{O_4}^{2-}(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[H_2P{O_4}^-] = 0.132 M - x$ $[HP{O_4}^{2-}] = 0.132M + x$ $[H_3O^+] = 0 + x$ 5. Calculate 'x' using the $K_a$ expression. $6.3\times 10^{- 8} = \frac{[HP{O_4}^{2-}][H_3O^+]}{[H_2P{O_4}^-]}$ $6.3\times 10^{- 8} = \frac{( 0.132 + x )* x}{ 0.132 - x}$ Considering 'x' has a very small value. $6.3\times 10^{- 8} = \frac{ 0.132 * x}{ 0.132}$ $6.3\times 10^{- 8} = 1x$ $\frac{ 6.3\times 10^{- 8}}{ 1} = x$ $x = 6.3\times 10^{- 8}$ Percent dissociation: $\frac{ 6.3\times 10^{- 8}}{ 0.132} \times 100\% = 4.77\times 10^{- 5}\%$ x = $[H_3O^+]$ $[H_2P{O_4}^-] = 0.132 M - x = 0.132 M - 6.3 \times 10^{-8}M \approx 0.132M$ $[HP{O_4}^{2-}] = 0.132M + x = 0.132 M + 6.3 \times 10^{-8}M \approx 0.132M$ $[H_3O^+] = 0 + x = 6.3 \times 10^{-8}M$ 6. Since we are adding a strong base, this reaction will occur: $H_2P{O_4}^-(aq) + OH^-(aq) -- \gt HP{O_4}^{2-}(aq) + H_2O(l)$ And these are the concentrations after this reaction: Remember: Reactants at equilibrium = Initial Concentration - y And Products = Initial Concentration + y Since $NaOH$ is a strong base, y = $[NaOH] = 5.31 \times 10^{-3}M$ $[H_2P{O_4}^-] = 0.132 M - 5.31 \times 10^{-3} = 0.127M$ $[HP{O_4}^{2-}] = 0.132M + 5.31 \times 10^{-3} = 0.137M$ 7. Now, calculate the hydronium ion concentration after the addition of the $NaOH$: $[H_3O^+] = Ka * (\frac{[H_2P{O_4}^-]}{[HP{O_4}^{2-}]})$ $[H_3O^+] = 6.3 \times 10^{-8} * \frac{0.127}{0.137}$ $[H_3O^+] = 6.3 \times 10^{-8} * 0.927$ $[H_3O^+] = 5.84 \times 10^{-8}$ 8. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 5.84 \times 10^{- 8})$ $pH = 7.23$