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Question

Calculate the pH of the resulting solution if 31.0 mL of 0.310 M HCl(aq) is added to 41.0 mL of 0.310 M NaOH(aq). pH=

Calculate the pH of the resulting solution if 31.0 mL of 0.310 M HCl(aq) is added to 21.0 mL of 0.410 M NaOH(aq). pH=

Answer 1

In the first part, the number of moles of HCl is 0.310 M/L x 1 L / 1000 mL x 31.0 mL = 0.00961 moles.

The number of moles of NaOH is 0.310 M/L x 1 L / 1000 mL x 41.0 mL = 0.0127 moles. After reaction there will be 0.0127 - 0.00961 = 0.00309 moles NaOH unreacted in a total volume of 72.0 mL. This calculates to a molarity of 0.00309 moles NaOH / 72.0 mL x 1000 mL = 0.0429 M. This means the OH^- concentration is also 0.0429 M.

This results in a pOH of - log [OH-] = 1.368. As pOH + pH = 14, the pH is 14 - 1.368 = 12.6 to 3 sig figs.

A similar approach can be used for the second part except that the HCl is in excess.

Please check math.