Calculate the pH of the resulting solution if 29.0 mL of 0.290 M HCl(aq) is added to
(a) 34.0 mL of 0.290 M NaOH(aq).
pH:?
(b) 39.0 mL of 0.340 M NaOH(aq).
pH:?
millimoles of HCl = 29 x 0.290 = 8.41
(a) 34.0 mL of 0.290 M NaOH(aq).
millimoles of NaOH = 34 x 0.290 = 9.86
base millimoles > acid millimoles
[OH-] = 9.86 - 8.41 /(34+29) = 0.023 M
pOH = -log[OH-] = -log (0.023) = 1.64
pH + pOH = 14
pH = 12.36
(b) 39.0 mL of 0.340 M NaOH(aq).
millimoles of NaOH = 39 x 0.340 = 13.26
base millimoles > acid millimoles
[OH-] = 13.26 - 8.41 /(39+29) = 0.0713 M
pOH = -log[OH-] = -log (0.0713) = 1.15
pH + pOH = 14
pH = 12.85
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