Calculate the pH of a #"0.2 M"# solution of #"NaC"_2"H"_3"O"_2# given that the hydrolysis constant is #K_h = 5.6xx10^(-10)#?

Question

Avery Adams · Accepted Answer

pH = 9.02

An ICE table can assist us in performing the computation. #color(white)(mmmmmmm)"A"^"-" + "H"_2"O" ⇌ "HA" + "OH"^"-"# #"I/mol·L"^"-1":color(white)(mll)0.2color(white)(mmmmmm)0color(white)(mmll)0# #"C/mol·L"^"-1":color(white)(mll)"-"xcolor(white)(mmmmml)"+"xcolor(white)(mm)"+"x# #"E/mol·L"^"-1":color(white)(m)"0.2-"xcolor(white)(mmmmml)xcolor(white)(mmll)x# #K_text(h) = (["HA"]["OH"^"-"])/(["A"^"-"]) = 5.6 × 10^"-10"# #x^2/("0.2-"x) = 5.6 × 10^"-10"# Utilize the five percent rule. #0.2/(5.6 × 10^"-10") = 3.6 × 10^8 > 400#. ∴ #x ≪0.2# Next #x^2/0.2 = 5.6 × 10^"-10"# #x^2 = 0.2(5.6 × 10^"-10") = 1.1 × 10^"-10"# #x = 1.1 × 10^"-5"# #["OH"^"-"] = x color(white)(l)"mol/L" = 1.1 × 10^"-5"color(white)(l) "mol/L"# #"pOH" = "-log"["OH"^"-"] = "-log"(1.1 × 10^"-5") = 4.98# #"pH = 14.00 - pOH = 14.00 - 4.98 = 9.02"#

Jeremiah Adams · Answer

undefined To calculate the pH of the given solution of $NaC_2H_3O_2$, first, find the concentration of $OH^-$ ions produced by the hydrolysis of the salt. Then, use the $K_w$ expression to find the concentration of $H^+$ ions. Finally, calculate the pH using the formula for pH:

$$pH = -\log[H^+]$$

Given:
$$K_h = 5.6 \times 10^{-10}$$
$$NaC_2H_3O_2 = 0.2 M$$

$$NaC_2H_3O_2 \rightarrow Na^+ + C_2H_3O_2^-$$

The hydrolysis of $C_2H_3O_2^-$ produces $OH^-$ ions. Let's denote the initial concentration of $C_2H_3O_2^-$ as $x$. Since $NaC_2H_3O_2$ dissociates completely, the concentration of $C_2H_3O_2^-$ ions is equal to the concentration of $OH^-$ ions produced.

$$K_h = \frac{[OH^-][CH_3COOH]}{[NaC_2H_3O_2]}$$

Solve for $[OH^-]$:

$$5.6 \times 10^{-10} = \frac{x^2}{0.2}$$

$$x = \sqrt{(5.6 \times 10^{-10}) \times 0.2}$$

$$x ≈ 1.49 \times 10^{-5} \, M$$

Now, we'll use $K_w$ to find the concentration of $H^+$ ions:

$$K_w = [H^+][OH^-]$$

$$1.0 \times 10^{-14} = (1.49 \times 10^{-5})[H^+]$$

$$ [H^+] ≈ \frac{1.0 \times 10^{-14}}{1.49 \times 10^{-5}}$$

$$ [H^+] ≈ 6.71 \times 10^{-10} \, M$$

Finally, calculate the pH:

$$pH = -\log(6.71 \times 10^{-10})$$

$$pH ≈ 9.17$$