Question

Calculate the ph of resulting solution when 50 ml of 0.2 molar hcl is mixed with 50 ml of 0.2 molar ch3oh h

Answer 1

As we know that

Mole = concentration x volume

So,

mole of NaOH = 50ml x 0.2 = 10.0

mole of HCl = 50ml x 0.4= 20.0

10 mole of NaOH exactly neutralized by 10 mole of HCl.

NaOH + HCl = NaCl + H2O

that is 20 -10 = 10 HCl mole

Con of H+ = no of mole of H+ / total volume of solution (50+50 ml = 100 ml)

= 10/100 = 0.1

pH = log ( H+)

= log (0.1) = 1

s0 the pH of solution will be one(1)

Answer 2

As we know that

Mole = concentration x volume

So,

mole of NaOH = 50ml x 0.2 = 10.0

mole of HCl = 50ml x 0.4= 20.0

10 mole of NaOH exactly neutralized by 10 mole of HCl.

NaOH + HCl = NaCl + H2O

that is 20 -10 = 10 HCl mole

Con of H+ = no of mole of H+ / total volume of solution (50+50 ml = 100 ml)

= 10/100 = 0.1

pH = log ( H+)

= log (0.1) = 1

s0 the pH of solution will be one(1)