Suppose you mix 25.0 mL 0.50 M Sr(OH)2 with 40.0 mL of 0.50 M HCl. Calculate the pH of the resulting solution.
[OH-] = 2*[Sr(OH)2]
= 2*0.50 M
= 1.0 M
[H+] = [HCl]
= 0.50 M
Given:
M(H+) = 0.5 M
V(H+) = 40 mL
M(OH-) = 1 M
V(OH-) = 25 mL
mol(H+) = M(H+) * V(H+)
mol(H+) = 0.5 M * 40 mL = 20 mmol
mol(OH-) = M(OH-) * V(OH-)
mol(OH-) = 1 M * 25 mL = 25 mmol
We have:
mol(H+) = 20 mmol
mol(OH-) = 25 mmol
20 mmol of both will react
remaining mol of OH- = 5 mmol
Total volume = 65.0 mL
[OH-]= mol of base remaining / volume
[OH-] = 5 mmol/65.0 mL
= 7.692*10^-2 M
use:
pOH = -log [OH-]
= -log (7.692*10^-2)
= 1.1139
use:
PH = 14 - pOH
= 14 - 1.1139
= 12.8861
Answer: 12.89
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