Question

calculate the mass of sodium hydroxide that must be added to 2.00 l of 1.00 m hc₂h₃o₂ to double the ph of the solution (assume that the added naoh does not change the volume of the solution). the kₐ for acetic acid is 1.8 × 10⁻⁵.
mass = g

Explanation:

Step1: Find initial \([H^+]\) from \(K_a\)

For a weak acid \(HC_2H_3O_2\) (acetic acid), \(K_a = 1.8\times10^{-5}\), initial concentration \(c = 1.00\ M\). Assume \(x\) is the dissociated \([H^+]\), then \(K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}\approx\frac{x^2}{c}\) (since \(x\ll c\)). So \(x=\sqrt{K_a\times c}=\sqrt{1.8\times10^{-5}\times1.00}\approx4.24\times10^{-3}\ M\). Initial \(pH_1 = -\log(4.24\times10^{-3})\approx2.37\).

Step2: Determine final \(pH\)

We need to double the \(pH\), so \(pH_2 = 2\times2.37 = 4.74\). Then \([H^+]_2 = 10^{-4.74}\approx1.82\times10^{-5}\ M\).

Step3: Reaction with \(NaOH\)

The reaction is \(HC_2H_3O_2 + NaOH
ightarrow NaC_2H_3O_2 + H_2O\). Let \(n_{NaOH}\) be moles of \(NaOH\) added. Initial moles of \(HC_2H_3O_2\) is \(n_0 = 1.00\ M\times2.00\ L = 2.00\ mol\). After adding \(NaOH\), moles of \(HC_2H_3O_2\) is \(n_0 - n_{NaOH}\), moles of \(C_2H_3O_2^-\) is \(n_{NaOH}\). Using Henderson - Hasselbalch equation: \(pH = pK_a+\log\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]}\). \(pK_a = -\log(1.8\times10^{-5})\approx4.74\). At \(pH = 4.74\), \(\log\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]} = 0\), so \([C_2H_3O_2^-]=[HC_2H_3O_2]\). Since moles are proportional to concentration (same volume), \(n_{NaOH}=n_0 - n_{NaOH}\), so \(2n_{NaOH}=n_0\), \(n_{NaOH}=\frac{n_0}{2}=\frac{2.00\ mol}{2}=1.00\ mol\).

Step4: Calculate mass of \(NaOH\)

Molar mass of \(NaOH\) is \(M = 40.00\ g/mol\). Mass \(m = n\times M=1.00\ mol\times40.00\ g/mol = 40.0\ g\). Wait, let's re - check the Henderson - Hasselbalch step. Wait, initial \(pH\) calculation: actually, for acetic acid, \(K_a = 1.8\times10^{-5}\), \(c = 1.00\ M\), the approximation \(x=\sqrt{K_a c}\) is valid. Initial \([H^+]=\sqrt{1.8e - 5\times1}=4.24e - 3\), \(pH = 2.37\), final \(pH = 4.74\). Using Henderson - Hasselbalch: \(pH = pK_a+\log\frac{[A^-]}{[HA]}\), \(pK_a = 4.74\), so when \(pH = 4.74\), \(\log\frac{[A^-]}{[HA]} = 0\), so \([A^-]=[HA]\). Moles of \(HA\) initially: \(2.00\ L\times1.00\ M = 2.00\ mol\). Let moles of \(NaOH\) be \(y\), then moles of \(HA\) after reaction: \(2.00 - y\), moles of \(A^-\): \(y\). So \(2.00 - y=y\), \(y = 1.00\ mol\). Mass of \(NaOH\): \(1.00\ mol\times40.00\ g/mol = 40.0\ g\). Wait, but let's check the \([H^+]\) approach again. The final \([H^+]\) is \(1.82e - 5\), which is much less than initial \([H^+]\), so the reaction with \(NaOH\) is necessary. The key is that to reach \(pH = 4.74\) (which is \(pK_a\)), the ratio of conjugate base to acid is 1, so half of the acid is neutralized. So moles of \(NaOH\) needed is half of initial moles of acid. Initial moles of acid: \(2.00\ L\times1.00\ M = 2.00\ mol\), so moles of \(NaOH\) is \(1.00\ mol\), mass is \(1.00\ mol\times40.0\ g/mol = 40.0\ g\).

Answer:

\(\boxed{40.0}\)