Stomach acid, a dilute solution of HCl can be neutralised by reaction with aluminig hydroxide
`Al (OH)_(3) + 3HCl (aq) to AlCl _(3) + 3H _(2)O`
How many milliliters of `0.1 M Al (OH) _(3)` solution are needed to neutralise 21 mL of `0.1 M HCl` ?

Options

A. 1) 14 mL
B. 2) 7 mL
C. 3) 21 mL
D. 4) none of these

Answer

B

Explanation

`M _(1) xx V _(1) = M_(2) xx V _(2)`
`because 0.1 M Al (OH) _(3)` gives `3 xx 0.1 =0.3 M OH ^(-)` ions
`0.3 xx V _(1) = 0.1 xx 21`
`V _(1) = (0.1 xx 21)/( 0.3) = 7 ml`