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VladimirAG [237]
3 years ago
10

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?

Chemistry
1 answer:
sattari [20]3 years ago
7 0
The main formulas are <span>pH=pKa + log(Base/Acid) and pKa = </span><span>-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  </span><span>log(Base/Acid)
for that the concentration of NaOH is [NaOH] = </span><span>500.0  x 0.110  / 500+535 =0.053M, the concentration of the Acid is  [Acid] =535*0.25 / </span><span>500+535 =0.12M, so its difference is </span><span><span>[Acid]-</span>[NaOH] = 0.12-0.053=0.07
so pH=</span><span><span>pKa + log(Base/Acid)= 9.66 + log(0.053 / 0.07)= 9.66-0.36=9.29</span> </span>
so pH=9.29.





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