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4 years ago

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?

1 answer:

7 0

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.

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