Question

A. A solution is made by combining 15.0 mL of 18.5 M acetic acid with 5.50 g of sodium acetate and diluting to a total volume of 1.50 L. Calculate the pH of the solution.

B.A 115.0 −mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 225.0 −mL sample of a solution that is 0.14 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

A: Moles of acetic acid(CH3COOH) taken = MxV = 18.5 mol/L x 0.015 L = 0.2775 mol

Moles of sodium acetate(CH3COONa) taken = mass of CH3COONa / molar mass of CH3COONa

= 5.50 g / 82.03 g/mol = 0.067 mol

Total volume of the buffer solution prepared, Vt = 1.50 L

Hence [CH3COOH] = moles of CH3COOH / Vt = 0.2775 mol / 1.50 L = 0.185 M

[CH3COONa] = moles of CH3COONa / Vt = 0.067 mol / 1.50 L = 0.0447 M

Now the pH of a buffer solution can be calculated from Hendersen equation

pH = pKa + log[CH3COONa] / [CH3COOH] = 4.75 + log (0.0447 M / 0.185M) = 4.13 (answer)

(B): Initial moles of AgNO3 = initial moles of Ag+(aq) = MxV = 3.0x10^-3 mol/L x 0.115 L = 0.000345 mol

Initial moles of NaCN = initial moles of CN-(aq) = MxV = 0.14 mol/L x 0.225 L = 0.0315 mol

Total volume after mixing, Vt = 0.115 L + 0.225 L = 0.340 L

Hence inital concentrations, [Ag+(aq)] = 0.000345 mol / 0.340 L = 0.00101M

[CN-(aq)] = 0.0315 mol / 0.340 L = 0.0926M

After mixing, Ag+(aq) and CN-(aq) will combine to form AgCN(s)

Ag+(aq) + CN-(aq) ---------> AgCN(s) ; Ksp = 1.6x10^-14

Ksp = 1.6x10^-14 = SxS

=> Solubility, S = underroot(1.6x10^-14) = 1.265x10^-7 M

Hence equilibrium concentration of Ag+(aq) = 0.001015M - 1.265x10^-7 M 0.00101M (answer)