Question

Calculate the pH change when 10. mL of 3.0 M NaOH is added to 500. mL of the following:

(a) pure water

(b) 0.10 M CH3COO-

(c)0.10 M CH3COOH

(d)a solution that is 0.10 M CH3COO- and 0.10 M CH3COOH

Answer 1

total volume after mixing = 500mL + 10 mL = 510 mL = 0.51 L

10mL of 3M NaOH = 0.03 moles

concentration of NaOH after mixing = 0.03/510 = 0.059 M

pOH = -log [0.059] = 1.23

pH = 14-pOH = 14-1.23 = 12.77

pH change = 12.77-7 = 5.77 (pure water has pH = 7)

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CH3COO- + H2O ----------> CH3COOH + OH-

CH3COO- will hydrolyze in water to form CH3COOH

moles of acetate = 0.1 *0.5 = 0.05 moles

CH3COO- is a strong base. So it will readily accept a proton from water.

moles of acetate = moles of acetic acid = moles OH- =0.05 moles

concentration of OH- = 0.05/0.5 = 0.1 M

pOH of the solution initially = 1. pH = 13

after adding NaOH, total concentration = 0.03 + 0.05 = 0.08

conentration = 0.08/0.51 = 0.157 M

pOH = -log[0.157] = 0.8. pH = 13.2

chanhe in pH = 13.2-13= 0.2

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CH3COOH -----> H+   + CH3COO

Since acetic acid is a weak acid it will ionize partially. Ka of acetic acid = 1.75 *10^-5

Ka = [H+][Ch3COO-]/[CH3COOH]

[H+] = [CH3COO-] =x

x^2 = 1.75 *10^-5 * 0.1

x = 1.32 *10^-6

pH = 5.88

Initial pH = 5.88

NaOH when added will form a buffer solution . [CH3COO-] = moles of NaOh added = 0.03

pH = pKa + log [0.03/0.1]

      = 4.23

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pH of the initial buffer

pH = pKa + log [0.1/0.1] = 4.75

NaOh will neutralize the acid and in the proces it will produce some acetate.

moles of NaOH added = 0.03

moles of acetate present = 0.1 *0.5 = 0.05

moles of acetic acid present = 0.1 80.5 = 0.05

Now the pH will be

pH = pka + log [(0.05 + 0.03)/(0.05-0.03)] = 5.18

pH change = 5.18-4.75 = 0.42