Calculate the pH of the solution when 50 ml 0.5 M Barium hydroxide react with 80 ml o.5 M sulphuric acid (H2So4)?

Q.Calculate the pH of the solution when 50 ml 0.5 M Barium hydroxide react with 80 ml o.5 M sulphuric acid (H2So4)?

Step 1:

The reaction between Ba(OH)2 and H2So4 is-

Ba (OH)2+ H2S04---->BaSo4 + H20

Now 50 mL of O.5 M Ba(oH)2 solution contain

=[(0.5 x 50 )÷1000] mole Ba(oH)2

=O. 025 mole Ba(oH)2

Again 80 mL 0.5M Sulphuric aud have

=[(.5×80)÷1000] mole

= 0.04 mole H2S04

Step 2:

According to reaetion
with 1 mole H2S041 mole Ba(o H)2 reaet
So, o.025 mole Ba (OH)2 Consume o.025 mole H2So4

:: ExCess amount of H2SO4
O.04 - 0.025
=0.015 mole H2So4

0.015 mole of  H2 S04 in unreacted
and present in soution.

Step 3:

we know,
PH, = - log [H+]
where
[H+] is themolar Concentration of H+ ion.

1 mole H2S04= 2 mole H+ ion

* 0.015 mole H2 S04 =2x 0.O15 mole H+ ion
=0.03 mole H+ ion,

Now total volume of solution
=(50+80) mL
= 130 ML

. Molar Concentration of H+ ion is-
=[(0.03×1000)÷130]
= 0.23 M

we know

PH = - log [H)

= - log (0.23)
PH = 0. 63
:: PH of the solution 0.63.

Step 1:

Step 2                                                                      

Step 3:                                                                 

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