Question

A student titrated a solution containing 3.7066 g of an unknown Diprotic acid to the end point using 28.94 ml of 0.3021 M KOH solution. What is the molar mass of the unknown acid? Hint: you must write a balanced equation for the reaction.​

Answer 1

Answer:

Molar mass of the diprotic acid is 424 grams

Explanation:

[hint: diprotic acid only contains 2 hydrogen protons]

Ionic equation:

${ \bf{2OH { }^{ - } _{(aq)} + 2H { }^{ + } _{(aq)}→ 2H _{2} O _{(l)} }}$

first, we get moles of potassium hydroxide in 28.94 ml :

${ \sf{1 \: l \: of \: KOH \: contains \: 0.3021 \: moles}} \\ { \sf{0.02894 \: l \: of \: KOH \: contain \: (0.02894 \times 0.3021) \: moles}} \\ { \underline{ = 0.008743 \: moles}}$

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.

Therefore, moles of acid that reacted are 0.008743 moles.

${ \sf{0.008743 \: moles \: of \: acid \: weigh \: 3.7066 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 3.7066}{0.008743}) \: g }} \\ = { \underline{423.95 \: g \approx424 \: grams}}$

for the molar mass: