Question

1.. A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 29.0 mL of NaOH.

2. A saturated solution of magnesium fluoride, MgF2, was prepared by dissolving solid MgF2 in water. The concentration of Mg2+ ion in the solution was found to be 1.18×10⁻³ M . Calculate Ksp for MgF2.

3. The value of Ksp for silver carbonate, Ag2CO3, is 8.10×10−12. Calculate the solubility of Ag2CO3 in grams per liter.

4. Lead thiocyanate, Pb(SCN)2, has a Ksp value of 2.00×10−5.. Calculate the molar solubility of lead thiocyanate in pure water. The molar solubility is the maximum amount of lead thiocyanate the solution can hold.

5. Calculate the molar solubility of lead thiocyanate in 1.00 M KSCN.

Answer 1

Millimole of CH3COOH in solution = molarity x volume in mL

= 0.35 M x 52 mL = 18.2 mmol

Millimole of NaOH in solution = 0.4 M x 29 mL = 11.6 mmol

Reaction is

CH3COOH ( aq, ) + NaOH ( aq ) $\rightarrow$CH3COONa ( aq) + H2O ( l )

So according to reaction 11.6 mmol of NaOH will react completely and same millimole of CH3COONa will formed and ( 18.2 - 11.6 = 6.6 mol ) acetic acid will remain.

Since CH3COOH + CH3COONa will form buffer whose p^{​​​​​H} is calculated by Henderson equation

P^{​​​​H} = p^{​​​​​ka} + log { [ CH3COONa] / [ CH3COOH ]}

P^{​​​​​​H =} - log ( 1.8 x 10^-5 ) + log{ ( 11.6 mmol / 81 mL ) / ( 18.2mol - 11.6 mol ) / 81 mL }

Since total volume = 52 + 29 = 81 mL

P^{​​​​​​H} = 4.97