Question

1. (I)- Calculate pH and pOH to 2 decimal places for the following:

a) 1.43×10^-1 M CaH₂

b) 8.77×10^-3 M KH

(II)- Determin pH&pOH to two decimal places for a solution that is produced by:

c) by mixing 883 mL of 6.18×10^-5 M KH with 251 mL of 4.86×10^-5 M KOH.

D) by mixing 79.4 mL of 5.66×10^-3 M HNO₃ with 5.87 mL of 2.11×10^-2 M Ca(OH)₂.

Answer 1

1. (I)- Calculate pH and pOH to 2 decimal places for the following:

a) 1.43×10^-1 M CaH₂

­Solution :- CaH2 is the strong base therefore

Concentration of the [OH-] = 1.43*10^-1 M CaH2 * 2 mol OH- / 1 mol CaH2 = 0.286 M

pOH = -log [OH-]

pOH= -log [0.286]

pOH= 0.544

pH= 14 – pOH

pH= 14 – 0.544

pH= 13.46

b) 8.77×10^-3 M KH

Solution :-

KH is strong base therefore [OH-] = 8.77*10^-3 M

pOH = -log [OH-]

pOH= -log [8.77*10^-3]

pOH= 2.06

pH= 14 – pOH

pH= 14 – 2.06

pH= 11.94

(II)- Determin pH&pOH to two decimal places for a solution that is produced by:

c) by mixing 883 mL of 6.18×10^-5 M KH with 251 mL of 4.86×10^-5 M KOH.

Solution :-

Both are strong bases

Therefore total moles of the OH- ions that can be formed are

Moles of KH = 6.18*10^-5 mol per L * 0.883 L = 5.46*10^-5 mol

Moles of KOH =4.86*10^-5 mol per L * 0.251 L = 1.22*10^-5 mol

Total moles = 5.46*10^-5 + 1.22*10^-5 =6.68*10^-5 mol

Total volume = 883 ml + 251 ml = 1134 ml = 1.134 L

Total concentration = 6.68*10^-5 mol / 1.134 L = 5.89*10^-5 M

pOH = -log [ OH-]

pOH = -log [5.89*10^-5]

pOH= 4.23

pH= 14 – 4.23

pH= 9.77

D) by mixing 79.4 mL of 5.66×10^-3 M HNO₃ with 5.87 mL of 2.11×10^-2 M Ca(OH)₂.

Solution :-

Moles of NHO3 = 5.66*10^-3 mol per L * 0.0794 L = 4.494*10^-4 mol

Moles of Ca(OH)2 = 2.11-10^-2 mol per L * 0.00587 L = 0.0001238 mol

Moles of HNO3 needed = 0.0001238 mol Ca(OH)2 * 2 mol HNO3 / 1 mol Ca(OH)2 = 0.0002477 mol HNO3

HNO3 is excess reagent

Therefore moles of HNO3 remain after reaction = 0.0004494 mol – 0.0002477 mol = 0.0002017 mol

Total volume = 79.4 ml + 5.87 ml = 85.27 ml = 0.08527 L

New molarity of the HNO3 = 0.0002017 mol / 0.08527 L = 0.002365 M

pH= -log [H+]

pH= -log [0.002365]

pH= 2.63

pOH = 14 – pH

pOH = 14 – 2.63

pOH = 11.37