Question

A 1.000 gram sample of K2CO3, (MM 138.2055g mol-1) is dissolved in enough water to make 250.00mL of solution. A 25.00mL aliquot is taken and titrated with 0.1000M HCl.
K2CO3(aq) + 2HCl(aq) ⇨ 2KCl(aq) + CO2(g) + H2O(l)
How many mL of HCl are used?

Answer 1

If 1g of K2CO3 is in 250ml of solution then, 25ml of the solution will contain ;

($25/250)*1$ =0.1g

If 1 mole of K2CO3 contain 138.2055g then 0.1g is equivalent to :

$(0.1/138.2055)*1$ = 0.00072moles

If 0.00072 mole of K2CO3 react then 0.00144moles of HCl reacted

If 0.1moles of HCl are in 1litre then,

$(0.00144/0.1) * 1000$ =14.4ml of 0.1000M HCl reacted