Question

1 ml of 1 mol/L HCL was diluted to 100 ml with water 10 ml of this solution was further diluted to 100 ml with water. what is ph of the final solution

Answer 1

pH = 0.05

IM = 1 mol / L

[ HCl ] = 1M

[ HCl ] = 1 mol/L

since...

1 ml = 0.001 L

[ HCl ] ' = 1 mol / ( 1 + 0.001 ) L.

again since....

10 ml = 0.01 L & 100 ml = 0.1 L

[ HCl ] " = 1 mol / ( 1 + 0.001 + 0.01 + 0.1 ) L

[ HCl ] " = 1 mol / 1.111 L

[ HCl ] " = 0.9 mol/ L

[ HCl ] " = 0.9 M

now....

pH = - log [ H+]

pH = - log [ 0.9 ]

pH = - × ( - 0.045 )

pH ~ 0.05