Question

500.0 Ml of 0.160 M NaOH is added to 565 ml of 0.250 M weak acid. Ka=2.46x10^-5. What is the pH of the resulting buffer
mmols HA = 565 x 0.25M = 141.25
mmols NaOH = 500 x 0.160 = 80

........HA + OH^- ==> A^- + H2O
I......141.25..0......0.......0
add..........80.................
C.......-80.-80.......+80
E........?....0.......80........

Substitute the E line into the Henderson-Hasselbalch equation and solve for pH.
pH = pKa + log (base/acid)

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