Question

28 mL of 0.10 M HCl is added to 60.0 mL of 0.10 M Sr(OH)2. Determine the concentration of OHin the resulting solution.

Answer 1

Answer:

Explanation:

Sr(OH)₂.+ 2HCl = SrCl₂ + 2H₂O

Moles of HCl in 28mL of .10 M HCl = .028 x .1 = .0028 moles .

Moles of Sr(OH)₂ in 60mL of .10 M Sr(OH)₂ = .060 x .1 = .0060 moles

2 moles of HCl  reacts with 1 mole of Sr(OH)₂

.0028 moles of HCl  reacts with .0014  mole of Sr(OH)₂

moles of Sr(OH)₂ remaining = .0060 - .0014 = .0046 moles .

Sr(OH)₂  = Sr⁺  +  2OH⁻

1 mole                  2 mole

.0046                    .0092

Total volume of solution = 88 mL .

88 mL of solution contains .0092 moles of OH⁻

concentration of OH⁻ = .0092 / .088

= .1045 M  .