Question

Calcuate the pH when 80 mL of 1.25 M HCL is added to 779 mL of 1.60 M of sodium acetate.

Please show two different ways hows to solve with steps and forumulas. Thanks!

Answer 1

Volume of HCl solution, V1 = 80 mL = 80 mLx(1L/1000mL) = 0.08L

Concentration of HCl solution, M1 = 1.25 M

Hence moles of HCl in the solution = M1V1 = 1.25Mx0.08L = 0.1 mol

Volume of CH3COONa solution, V2 = 779 mL = 779mLx(1L/1000mL) = 0.779L

Concentration of CH3COONa solution, M2 = 1.60 M

Hence moles of CH3COONa in the solution = M2V2 = 1.60Mx0.779L = 1.2464 mol

When HCl and CH3COONa are added they react with each other to form weak acid CH3COOH and NaCl. Since HCl is present in smaller amount (0.1 mol) in comparison to CH3COONa (1.2464 mol), HCl will be completely used up and acts as limiting reagent.

HCl + CH3COONa --------------->CH3COOH + NaCl

1 mol 1 mol 1 mol

(0.1 - 0.1)mol (1.2464 - 0.1)mol 0.1 mol

= 0 1.1464 mol 0.1 mol

Hence 0.1 mol of HCl completely reacts with 0.1 mole of CH3COONa to form 0.1 mol of CH3COOH

Total volume of solution after mixng, Vt = 0.08L+0.779L = 0.859 L

After the reaction, [CH3COONa] = 1.1464 mol / Vt = 1.1464mol / 0.859 L = 1.335 M

[CH3COOH] = 0.1 mol / Vt = 0.1mol / 0.859L = 0.1164 M

Since CH3COOH is a weak acid and CH3COONa is a salt of CH3COOH with a strong base NaOH, hence the above solution after mixing will act as a buffer solution.

Now we can calculate the pH of this buffer solution by applying Henderson's equation which is

pH = pKa + log [salt] / [acid]

pKa value for CH3COOH = 4.76

[salt] = [CH3COONa] = 1.335 M

[acid] = [CH3COOH] = 0.1164 M

Hence pH =  pKa + log [salt] / [acid] = 4.76 + log (1.335 M / 0.1164 M) = 5.82 (answer)