Question

the pH of a solution obtained by mixing 50 ml of 0.4 N NH4Cl and 50 ml of 0.2 N NaOH is ​

Answer 1

Mole = concentration x volume

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

hi there

pH of the solution can be very easily calculated if we know the normality of the solution. So, here 2 solutions are mixed. If we calculate the normality of resulting solution, the question is solved. Here are the steps:

No. of miliequivalent of acid=50*0.4=20

No. of miliequivalent of base=50*0.2=10

Now, as the

miliequivalent of acid>miliequivalent of base

Remaining miliequivalent of acid=10 which is also equal to that of H+.

Normality of H+ =10/(50+50) = 0.1

So, pH =-log(N)=1

hope it helps