Look at the equation. Base + acid = salt + water.
NaOH + HC2H3O2 ==> NaC2H3O2 + H2O
Calculate the moles NaOH to start.
Calculate the moles HC2H3O2 to start. I haven't done the calculation but I assume that the HC2H3O2 will be in excess and all of the NaOH will be used. So you end up with a buffered solution, which in this case is a weak acid (HC2H3O2) and its salt (in this case NaC2H3O2.
Now use the Henderson-Hasselbalch equation,
pH = pKa + log (base/acid) to calculate pH.
What is the pH of the solution created by combining 0.80 mL of the 0.10 M NaOH(aq)with 8.00 mL of the 0.10 M HC2H3O2(aq)?
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1 answer