Question

How would we work this out? I know the answer is 4.89

44) In the titration of 10.00mL of a 0.15M acetic acid solution (Ka 1.8 x 10-5) with 0.25M sodium hydroxide, what is the pH after 3.50 mL of the base has been added? a) 4.59 b) 4.69 c) 4.79 d) 4.89 e) 4.99

Answer 1

Write the reaction between acetic acid and NaOH as follows: CH,COOH(aq) + NaOH(aq) - CH2COONa (aq) + H20(1) PK = -log K = -log(1.8x10-S) = 4.74 volume of CH2COOH = 10.00 mL molarity of CH,COOH = 0.15 M moles of CH2COOH = 0.15 MM 1L MX10.00 mLx 1000 ml = 0.0015 mol volume of NaOH = 3.50 mL molarity of NaOH = 0.25M moles of NaOH = 0.25M xmo * M 1L -x 3.50 mx- *2.20 1000 mL F = 0.000875 mol moles of CH COOH remaining = 0.0015 mol - 0.000875 mol = 0.000625 mol moles of CH,COONa formed = 0.000875 mol ---------------------------------------- -- - Write the relation between pH and pK, as follows: pH = pK, +logCH,COONa] [CH,COOH) moles of CH,COONa pH = pK, +log moles of CH,COOH Substitute the correspodning values in the above equation and calculate pH. (0.000875 pH = 4.74+log 0.000625) pH = 4.74+log(1.4) pH = 4.74+0.15 pH = 4.89 Therefore, the required solution is 4.89)