If 30 mL of 0.10 M NaOH is added to 50 mL of 0.20 M HC 2 H 3 O 2 , what is the p

Question

1. If 30 mL of 0.10 M NaOH is added to 50 mL of 0.20 M HC2H3O2, what is the pH of the resulting solution at 25°C?  Ka for HC2H3O2is 1.8 × 10-5 at 25°C. A. 2.7 B. 5.1 C. 4.2 D. 4.4 E. 10.1

1. If 30 mL of 0.10 M NaOH is added to 50 mL of 0.20 M HC2H3O2, what is the pH of the resulting solution at 25°C?  Ka for HC2H3O2is 1.8 × 10-5 at 25°C. A. 2.7 B. 5.1 C. 4.2 D. 4.4 E. 10.1

If 30 mL of 0.10 M NaOH is added to 50 mL of 0.20 M HC2H3O2, what is the pH of the resulting solution at 25°C?  Ka for HC2H3O2is 1.8 × 10-5 at 25°C. A. 2.7 B. 5.1 C. 4.2 D. 4.4 E. 10.1 If 30 mL of 0.10 M NaOH is added to 50 mL of 0.20 M HC2H3O2, what is the pH of the resulting solution at 25°C?  Ka for HC2H3O2is 1.8 × 10-5 at 25°C. 2.7 5.1 4.2 4.4 10.1 A. 2.7 B. 5.1 C. 4.2 D. 4.4 E. 10.1

Explanation / Answer

number of moles of NaOH = M*V = 0.1 M*0.03 L = 0.003 mol
number of moles of CH3COOH = M*V = 0.2 M*0.05 L = 0.01 mol
0.003 mol of each will react to form 0.003 mol of CH3COONa
So,
mol of CH3COONa formed = 0.003 mol
mol of CH3COOH left = 0.01 - 0.003 = 0.007 mol
Total volume = 30 mL + 50 mL = 80 mL = 0.08 L

[CH3COONa ] =Number of moles / volume = 0.003/0.08 = 0.0375 M
[CH3COOH ] =Number of moles / volume = 0.007/0.08 = 0.0875 M

Reaction taking place is:
CH3COOH + NaOH ---> CH3COONa + H2O

pKa = -log Ka = -log (1.8*10^-5) = 4.7447

use:
pH = pKa + log {[CH3COONa]/ [CH3COOH]}
       = 4.7447 + log (0.0375/0.0875)
       =4.4
Answer: 4.4

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