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Question
Calculate the PH of a solution formed by mixing equal volume of two solution A and B of a strong acid having PH=6 and Ph=4 respectively?
Calculate the PH of a solution formed by mixing equal volume of two solution A and B of a strong acid having PH=6 and Ph=4 respectively?
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Solution
We know pH = - log[H+]
For solution A
pH = 6 = -log[H+]
or log[H+]= -6
or [H+]= 10-6
Similarly for Acid B, [H+]=10-4
If initial volume of each acid was V then final volume is 2V.
Now we know V1S1=V2S2
Where V1 an S1 are initial volume and strength of acid, and V2 an S2 are final volume and strength of acid
So V1 = V
V2= 2V
V X 10-6= 2V X S2
S2= 10-6/2
Similarly for acid B final strength is 10-4/2
Total concentration after mixing (10-6/2+10-4/2)= 10-6/2(1+100) = 101X10-6/2
So pH= -log[101X10-6/2] = 4.30
So pH of resulting mixture is 4.30
For solution A
pH = 6 = -log[H+]
or log[H+]= -6
or [H+]= 10-6
Similarly for Acid B, [H+]=10-4
If initial volume of each acid was V then final volume is 2V.
Now we know V1S1=V2S2
Where V1 an S1 are initial volume and strength of acid, and V2 an S2 are final volume and strength of acid
So V1 = V
V2= 2V
V X 10-6= 2V X S2
S2= 10-6/2
Similarly for acid B final strength is 10-4/2
Total concentration after mixing (10-6/2+10-4/2)= 10-6/2(1+100) = 101X10-6/2
So pH= -log[101X10-6/2] = 4.30
So pH of resulting mixture is 4.30
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