500.0 ml of 0.110 m naoh is added to 535 ml of 0.250 m weak acid (ka = 6.37 * 10-5. what is the ph of the resulting buffer?

Question

Jacob Glass

Chemistry

8 May, 18:49

500.0 ml of 0.110 m naoh is added to 535 ml of 0.250 m weak acid (ka = 6.37 * 10-5. what is the ph of the resulting buffer?

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Alma Gonzales · Answer 1

The main formulas are pH=pKa + log (Base/Acid) and pKa = - log (Ka)

so firstly, we must find the value of pKa,

Ka=6.37 x 10 ^-5, and then logKa = log (6.37. 10^-5) = - 9.66, so - logKa = + 9.66=Ka

next let's find log (Base/Acid)

for that the concentration of NaOH is [NaOH] = 500.0 x 0.110 / 500+535 = 0.053M, the concentration of the Acid is [Acid] = 535*0.25 / 500+535 = 0.12M, so its difference is [Acid] - [NaOH] = 0.12-0.053=0.07

so pH = pKa + log (Base/Acid) = 9.66 + log (0.053 / 0.07) = 9.66-0.36=9.29

so pH=9.29.