Quick pH calculation - Please verify

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The discussion centers on calculating the new pH after adding sulfuric acid to a sodium hydroxide solution. Initially, the sodium hydroxide solution has a pH of 12.50, and upon adding 36.00 mL of 0.0200 mol/L sulfuric acid, the calculations show that the limiting reagent is sulfuric acid, which is fully consumed. After the reaction, 0.00014 mol of hydroxide ions remain, leading to a total solution volume of 86 mL. The final calculated pH of the resulting solution is 11.21, confirming the accuracy of the calculations. The discussion concludes with a request for verification of the method used to arrive at the pH.
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Quick pH calculation -- Please verify

A 50.0 mL aqueous solution of sodium hydroxide has a pH of 12.50. If 36.00 ml of 0.0200 mol/L sulfuric acid is added to this sodium hydroxide solution, what will be the new pH of the resulting solution? Assume that the temperature stays constant at 25C, and that the volumes are perfectly additive.

50ml=0.05L
0.00158 mol OH- initially.
0.00072 mol H2SO4
Entire consumption of H2SO4(limiting reagent).
0.00086 mol OH- left.
0.00072 mol HSO4- left.
Entire consumption of HSO4-:
0.00014 mol OH- left.
Volume total=86 ml.
ph=11.21


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