Question

35.0 mL sample of 0.150 M acetic acid (HC2H3O2) (Ka = 1.8 x 10-5) is titrated with 0.150 M NaOH solution. Calculate the pH after 17.5 mL volumes of base have been added

millimoles HAc = mL x M = 35.0 x 0.150 = approx 5.25
mmols NaOH = 17.5 x 0.150 = 2.625

..........HAc + NaOH ==> NaAc + H2O
I........5.25....0........0......0
add.............2.625...............
C.......-2.625..-2.625...+2.625..+2.625
E.......2.625.....0.....2.625.....2.625

Substitute the E line into the Henderson-Hasselbalch equation and solve for pH.

Related Questions

A 10.0 mL of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with .5062M NaOH, an... Acetic acid (HC2H3O2) is an important component of vinegar. A 10.00mL sample of vinegar is titrated... a 50ml aliquot sample of vinegar is titrated with a 0.078 M NaOH sol'n. The phenolpthalein end point... 10 ml sample of vinegar an aqueous solution 0f acetic acid( HC2H3O2) is titrated with 0.5062 M an...