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500.0 mL of 0.140 M NaOH is added to 565 mL of 0.250 M weak...

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Chemistry

500.0 mL of 0.140 M NaOH is added to 565 mL of 0.250 M weak acid(Ka = 8.39 × 10-5). What is the pH of the resulting buffer?

HA(aq)+OH^-(aq)=H2O(l)+ A^-(aq)

pH=?

Answer & Explanation Solved by verified expert
4.4 Ratings (752 Votes)
no of moles of NaOH 014 x 500 1000 007mol New conc of NaOH after addition to acid 007 1065 x 1000 00657 M No of    See Answer
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