What is total volume of solution(in ml) when 1M NaOH is required to be added in 100mL 1M CH3COOH(Ka=10-5) solution so that pH is 6.
1Here , its reaction between a strong acid and a weak base.
so,use pH = pKa + log [salt]/[acid]
here, [salt] cn be found out in M. it would perhaps come 10M.
now, volume of the acid and its molarity is known,along with base's molarity.
so,use: M1V1=M2V2
volume of base = 0.1 L.
1357let the volume of base added is V ml
mmol of salt formed=V
mmol of acid remined=100-V
we have [salt]/[acid]=10
V/(100-V)=10
or V=1000/11
total volume=V+100=2100/11 ml
1champ, here i found out the vol. of base = 0.1 L
So, total vol. of solution = 200 ml.