# question (1 Viewer)

#### Chang

##### Member
Hi guys can i get help with this question:

In some laboratory analyses the preferred technique is to dissolve an impure solid sample of acid or base in solution containing an excess of acid or base, and then 'back-titrate' the excess with a standard base or acid of known concentration.

This technique is used to assess the purity of a sample of KHSO4.
A 0.4696-g solid sample of impure KHSO4 was dissolved in 50 mL of 0.1 M KOH. This solution contains more base than is needed to neutralise the acid, HSO4-. After neutralisation of the HSO4-, the excess base is neutralised by 10.44 mL of 0.2127 M HCl.
Calculate the percentage by mass of pure KHSO4 in the sample.

express concerntrations of KHSO4 as %w/w

If you could working out would be appreciated, thx

#### Jumbo Cactuar

##### Argentous Fingers
amount KOH initially = 0.05*0.1 = 0.005 moles
amount KOH after HSO4- = 0.01044 * 0.2127 = 0.00222 moles
amount KHSO4 = 0.0028 moles
mass KHSO4 = 0.0028 * 136.178 = 0.3785g
concentration (w/w) = 0.3785/0.4696 *100 = 80.6%

#### Chang

##### Member
thx alot for the help 