Question

Find the pH of the following solution

10.0mL of solution A (which is .15 of .5M of acetic acid and .15 of .5M of sodium acetate) and 10.0mL of 0.1M of HNO3

10.0mL of solution A (same as above) and 10.0mL of 0.1M of NaOH

Answer 1

1) CH₃COONa + HNO₃ -----> CH₃COOH + NaNO₃

Number of moles of CH₃COOH = volume x molarity = 0.15 L x 0.5 M = 0.075 moles

Number of moles of CH₃COONa = volume x molarity = 0.15 L x 0.5 M = 0.075 moles

Number of moles of HNO₃ = volume x molarity = 0.01 L x 0.1 M = 0.001 moles

Now, 0.001 moles of HNO₃ will form 0.001 moles of CH₃COOH

Total moles of CH₃COOH in the solution now = 0.075 + 0.001 = 0.076 moles

Total volume of solution now = 0.15 + 0.15 + 0.01 = 0.31 L

Molarity of CH₃COOH now = 0.076 / 0.31 = 0.25 M

K_a of [CH3COOH] = 1.8 x 10^-5

Ka = [H⁺] [CH3COO^-] / [CH3COOH]

1.8 x 10^-5 = x²/0.25

x = 0.0021 M

pH = -log[H⁺]

pH = -log(0.0021) = 2.67

2) CH₃COOH + NaOH -----> CH₃COONa + H₂O

Number of moles of CH₃COOH = volume x molarity = 0.15 L x 0.5 M = 0.075 moles

Number of moles of CH₃COONa = volume x molarity = 0.15 L x 0.5 M = 0.075 moles

Number of moles of NaOH = volume x molarity = 0.01 L x 0.1 M = 0.001 moles

Now, 0.001 moles of HNO₃ will form 0.001 moles of CH₃COOH

Total moles of CH₃COONa in the solution now = 0.075 + 0.001 = 0.076 moles

Total volume of solution now = 0.15 + 0.15 + 0.01 = 0.31 L

Molarity of CH₃COONa now = 0.076 / 0.31 = 0.25 M

K_a of [CH3COOH] = 1.8 x 10^-5

Ka = [H⁺] [CH3COO^-] / [CH3COOH]

1.8 x 10^-5 = x(0.075+x)/(0.25-x)                [assuming x is very small]

x = 6 x 10^-5

pH = -log[H⁺]

pH = -log(6 x 10^-5) = 4.22