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A 0.30 M solution of an acid with KA= 3.0 x 10-7 is titrated with a...

A 0.30 M solution of an acid with KA= 3.0 x 10-7 is titrated with a 0.15 M solution of sodium hydroxide. Calculate the pH of the solution after 110 mL of the sodium hydroxide solution is added to 55 mL of the acid solution.

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Answer #1

for HA: 0.30m, 55 mL i. moles of UA = 0.30 x 55 = 16.5 mmol for Naol :0.15m, nom moles of NaOH = 0.15 x 10 = 16.5 mmol Here,Now, here ka of HA = 3x10 + . we know, ka koko ..Ko = 22 0 0.3 x 108 = x² ) x = 0.57810-4 .. [on] = 0.57 6104 pou - Llog con

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