Answer to Question #106980 in General Chemistry for Kyonnie

Question

An aqueous mixture of 0.5 mole CH3COOH and 0.5 mole of CH3COO– had 0.1 moles of NaOH added, bringing the total volume of the solution to 1.0 L at 25oC. What is the final pH of the solution, after the addition of HCl, if the Ka(CH3COOH) =1.8 x 10–5?

Answer 1

$$CH_3COOH + NaOH = CH_3COONa + H_2O$$

$$n(NaOH) = n(CH_3COOH) = n(CH_3COONa) = 0.1 mol$$

$$n(CH_3COOH)_{last} = 0.5mol - 0.1mol = 0.4 mol$$

$$n(CH_3COO^-) = 0.5mol+0.1 mol = 0.6 mol$$

$$pH =- log_{10}K_a + log_{10}([CH_3COOH]/[CH_3COO^-])$$

$$pH = -log_{10}(1.8*10^{-5}) + log_{10}(0.4/0.6)$$

$$pH = 4.74-0.17 = 4.57$$

If hydrochloric acid is further added, it is necessary to know its volume and concentration. Look carefully at the condition, maybe the pH of the solution was asked after adding the base, or check the missing data about the acid. Thanks!