Calculate the pH of a solution that is 2.00 M HF,1.00 M NaOH,and 0.642 M NaF.(K_a = 7.2 $\times$ 10^-⁴)

Question

Quizplus · Accepted Answer

First we need to write the balanced equation for the reaction between HF and NaOH:

HF + NaOH → NaF + H2O

Since NaOH is a strong base, it will completely dissociate to form Na+ and OH- ions:

NaOH → Na+ + OH-

We can use the initial concentrations of HF and NaOH, and the equilibrium concentration of NaF to calculate the concentrations of H+ and F- ions, and therefore the pH of the solution.

Step 1: Calculate the moles of NaOH:

moles NaOH = (1.00 M) x (1.00 L) = 1.00 mol

Step 2: Calculate the moles of HF that reacts with NaOH:

1 mol NaOH reacts with 1 mol HF, so:

moles HF reacted = 1.00 mol

The remaining moles of HF are:

moles HF remaining = (2.00 M) x (1.00 L) - 1.00 mol = 1.00 mol

Step 3: Calculate the moles of NaF:

moles NaF = (0.642 M) x (1.00 L) = 0.642 mol

Step 4: Calculate the equilibrium concentrations of H+ and F- ions:

[H+] = [OH-] = (1.00 mol + 0.642 mol) / 2.00 L = 0.821 M

[F-] = 0.642 mol / 2.00 L = 0.321 M

Step 5: Calculate the pH of the solution:

pH = -log[H+] = -log(0.821) = 0.087 + 0.917 = 3.36

Therefore, the answer is B) 3.36.

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Calculate the PH of a Solution That Is 2 $\times$ 10^-⁴)
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