Question

# determine the pH of a solution containing 15.00mL of 0.1163 M LiOH with 42.92mL of 0.05643M...

determine the pH of a solution containing 15.00mL of 0.1163 M LiOH with 42.92mL of 0.05643M HNO3

LiOH + HNO3 -----------> LiNO3 + H2O

moles of LiOH = 0.1163 x 15/1000 = 0.00174

moles of HNO3 = 0.05643 x 42.92 / 1000 = 0.002422

0.002422 - 0.00174 = 0.000682 moles HNO3 left

[HNO3] = 0.000682 / 0.05792

[HNO3] = 0.012 M

HNO3 is strong acid so [HNO3] = [H+] = 0.012 M

pH = - log [H+]

pH = - log [0.012]

pH = 1.92

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