Chem4 - Acids and Bases

Question 0

How to calculate the pH of a solution of a strong acid H2SO4 and a strong base ie NaOH and strong acid is in excess (6)

Answer 0

~work out moles of H2SO4
~work out moles of NaOH
~multiply moles of H2SO4 for moles of H+ (its diprotic, as double H+ molecules to moles of H2SO4 as H2 in molecule)
~take moles of OH- (same as NaOH) away from moles of H+ for excess moles of H+
~divide by new volume for conc H+ c=n/v (new vol = vol NaOH + vol H2SO4)
~plug into pH=-log[H+]

Question 1

Suggest why chloroethanoic acid is a stronger acid than ethanoic acid? (2)

Answer 1

~Cl )is more electronegative so) withdraws electrons

~weakens O-H bond or O-H bond is more polar

Question 2

How to calculate the pH of an equimolar solution (or at half neutralisation) of weak acid and its salt (4)

Answer 2

~when equimolar/half neutralised pH=pKa
~¥ [H+]=Ka (pH, -log[H+]= -log(Ka) ,cancel logs
~so -log(Ka value of weak acid)
~calculate, this is pH

Question 3

Suggest an alternative name for an acidity regulator, explain how it regulates acidity (3)

Answer 3

~buffer
~added H+ is removed by A- , H+ + A- –> HA
~¥ ratio of [HA]/[A-] stays almost constant