Use the henderson-hasselbalch equation
pH=pKa+log[A-/HA]
where
Pka=-logKa
Ka=1.3 x 10–5
HA=propanoic acid
A-=conjugate base of propanic acid
0.0300L *(0.165M)=moles of propanoic acid
0.010*(0.300M)= moles of KOH
0.005*(0.300M)= moles of KOH
Solve for moles of conjugate base of propanoic acid.
moles of propanoic acid-moles of KOH=moles of conjugate base of propanoic acid
moles of propanoic acid-moles of conjugate base of propanoic acid=moles of propanoic acid
***For the first problem, total volume is 40mL or 0.040 L
moles of propanoic acid/total volume=HA
moles of conjugate base of propanoic acid/total volume=A-
Substitute into the henderson hasselbalch equation and solve.
A 30.0-ml sample of 0.165 M propanoic acid is titrated with 0.300 M KOH. What is the pH at 5 ml of added base, at 10 ml of added base?
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