Question

A 0.2722 g sample of a pure carbonate, X n CO 3 ( s ) , was dissolved in 50.0 mL of 0.1200 M HCl ( aq ) . The excess HCl ( aq ) was back titrated with 23.60 mL of 0.0980 M NaOH ( aq ) . How many moles of HCl react with the carbonate?

Answer 1

0.00369 moles of HCl react with carbonate.

Step-by-step explanation:

Number of moles of HCl present initially =
`(0.1200)/(1000)* 50.0` moles = 0.00600 moles

Neutralization reaction (back titration):
`NaOH+HCl\rightarrow NaCl+H_(2)O`

According to above equation, 1 mol of NaOH reacts with 1 mol of 1 mol of HCl.

So, excess number of moles of HCl present = number of NaOH added for back titration =
`(0.0980)/(1000)* 23.60` moles = 0.00231 moles

So, mole of HCl reacts with carbonate = (Number of moles of HCl present initially) - (excess number of moles of HCl present) = (0.00600 - 0.00231) moles = 0.00369 moles

Hence, 0.00369 moles of HCl react with carbonate.