Question

A 0.30 M solution of an acid with K_A= 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.

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, Inaon] = [HA ] - This is equivalente point wondisson. HA + NaOH Na A + H₂O (e) 1 16.5 16.5 -16.5 -105 + 16.5 pod uw 16.5 = 16.5 mmol 165 m2 0.1 M [A-] = [NDA] - hotel moles Total volume I + H2O HA & on 1 0.1 c E -X 0.1-2

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 ) = -log (0.57 X104) IPOH = 4.24 Now, pH = 14 - 4.24 pH = 9.76