A solution of NaOH is prepared by dissolving 1.5 g of the base in 500 mL of water. Calculate the pH of the solution.

Explanation

The molarity M of the solution is given by
` " " w= (M M. V)/(1000)`
In the present case, w =1 .5. g M .(molecular mass) `= 23 + 16 + 1 = 40 and V - 500 ml.`
` therefore " " M= ( 1000w)/( M. V) = (1000 xx 1.5)/( 40 xx 500) = 0.075`
Thus, the concentration of the solution is` 0.075" mol " L^(-1) ` Since NaOH is a strong base,
` [OH^(-) ] =[NaOH] =0.075 " mol " L^(-1)`
` " " [H_3O^(+) ] =(K_w)/([OH^(-)])=(1.0 xx 10 ^(-14))/( 0.075)`
` " " =1.33 xx 10 ^(-13)`
Hence, ` " " pH = - log _(10 ) [H_3O^(+) ]`
`" "= -log _(10) (1.33 xx 1 0^(-13))`
` " " =12.88`