Question

Calculate the ph of a solution which contains 9.9 ml of 1m hcl and hundred ml of 0.1 n a o h

Answer 1

800 ml.of 0.05M NaOH contain

=800 x 0.05 =40 milli mole of NaOH

200 ml.of 0.1M HCl contains

= 200 X 0.1= 20 milli moles of HCl

So 20 milli moles of NaOH will react with 20 milli moles of HCl and the final solution of (800+200) 1000 ml.will contain 40–20=20 milli moles of NaOH.

So finally 1000 ml.of solution contains 20 milli moles of NaOH.

Ie., the concentration of the final solution is

=20/1000 =0.02N

So pOH of 0.02N NaOH =-log(2 x10-2)

=2–0.3010 =1.6990

So pH =14-pOH =14- 1.6990= 12.3010

Answer 2

HOLA MATE..!!

ANSWER:-

800 ml.of 0.05M NaOH contain

=800 x 0.05 =40 milli mole of NaOH

200 ml.of 0.1M HCl contains

= 200 X 0.1= 20 milli moles of HCl

So 20 milli moles of NaOH will react with 20 milli moles of HCl and the final solution of (800+200) 1000 ml.will contain 40–20=20 milli moles of NaOH.

So finally 1000 ml.of solution contains 20 milli moles of NaOH.

Ie., the concentration of the final solution is

=20/1000 =0.02N

So pOH of 0.02N NaOH =-log(2 x10-2)

=2–0.3010 =1.6990

So pH =14-pOH =14- 1.6990= 12.3010

HOPE IT HELPS..!!