If 100-mL of 1.0 M Sr(OH)2 is added to 100 mL of 1.0 M HCl, the pH of the mixture would be . Group of answer choices

pH = 13.7

Explanation:

A strong acid (HCl) reacts with a strong base Sr(OH)₂ producing water and a salt, thus:

2HCl + Sr(OH)₂ → 2H₂O + SrCl₂

To solve this problem, we need to find initial moles of both reactants and, with the chemical equation find limiting reactant and moles in excess to find pH as follows:

The initial moles of HCl and Sr(OH)₂ are:

100mL = 0.1L ₓ (1.0mol / L) = 0.100 moles of both HCl and Sr(OH)₂

As 2 moles of HCl reacts per mole of Sr(OH)₂, moles of Sr(OH)₂ that reacts with 0.100 moles of HCl are:

0.100 moles HCl ₓ (1 mol Sr(OH)₂ / 2 mol HCl) = 0.050 moles Sr(OH)₂

That means HCl is limiting reactant and after reaction will remain in solution:

0.100 mol - 0.050mol =

0.050 moles of Sr(OH)₂

Find pH:

1 mole of Sr(OH)₂ contains 2 moles of OH⁻, 0.050 moles contains 0.050×2 = 0.100 moles of OH⁻. In 200mL = 0.2L:, molar concentration of OH⁻ is:

0.100 moles / 0.2L =

[OH⁻] = 0.5M

As pOH of a solution is -log[OH⁻],

pOH = -log 0.5M

pOH = 0.301

And knowing:

pH = 14 - pOH

pH = 14 - 0.301

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