A sample of 3.95g of 60% pure chalk (impurities in the chalk are insoluble in HCl), is dissolved in 250 mL of 0.2(M) HCl. What volume `(cm^(3))` of 0.01 (N) NaOH is required to neutralise the excess acid?

Explanation

Amount of `CaCO_(3)=(60xx3.95)/(100)=2.37g`
`2.37g" "CaCO_(3)-=(2.37)/(100)-=2.37xx10^(-2)mol" "CaCO_(3)`
`CaCO_(3)+2HCl to CaCl_(2)+H_(2)O+CO_(2)`
`therefore` 1000 mL `2(M)HCL-=1molCaCO_(3)`
`2.37xx10^(-2)molCaCO_(3)-=(1000xx2.37xx10^(-2))mL2(M)HCl(l)`
`-=23.7mL2(m)HCl -=237mL" "0.2(M)HCl`
`therefore`Amount of HCl remained `=(250-237)mL" "0.2(M)HCl`
`=13mL0.2(M)HCl`
If V ml of 0.01 (N) or 0.01 (M) NaOH solution is required to neutralise the excess acid, then
`13xx0.2=Vxx0.01[V_(1)S_(1)=V_(2)S_(2)]" "therefore V=260mL`