24From HA : [H+]=√KaC=10-2
From HB : [H+]=√KaC=2*10-2
Total [H+]=10-2+2*10-22=3*10-22
pH=-log[H+]=2-log3+log2≈2-0.48+0.3≈1.82
Equal volumes of acid, 1M HA and acid 1M HB are mixed. Ka for acids are respectively, 10-4 and 4*10-4. Find the pH of the resultant solution.
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12 Answers
24
1357hmm.....
there's a little mistake there
Don't you think there will be a liittle less concentration of [H+] due to common ion effect.
24but due think it will matter a lot ???
answer will still remain abt 1.8
1357No it will not, but this question was given to solve by the traditional method only.
Let's do it here.
I m giving two methods here. Just use one which you like.
Method 1:
HA→H++A-
1 10-2 10-2
HB→H+ + B-
1 2x10-2 2x10-2
Initially these are the conditions for the acods where [H+]=√(KaC)
After adding them:
HA→H++A-
1/2 10-2/2 10-2/2
HB→H+ + B-
1/2 10-2 10-2
But total [H+] becomes 1.5x10-2
So there will be backward reaction in both the cases
HA→H++A-
1/2+x 1.5x10-2-x-y 10-2/2 -x
HB→H+ + B-
1/2+y 1.5x10-2-x-y 10-2-y
Solve for x and y from Ka's
Method 2:
K1=[H+][A-]/[HA], K2=[H+][B-]/[HB]
we have [HA]=[HB]=0.5
So there are three unknowns(H,A,B) and two equations.
third equation you can use here is balance of charge:
So we have [H+]=[A-]+[B-]
Now solve and find [H+]
24kk
but if this ques comes in exam..my method will work fine naa???unless they want ans correct to second place of decimal..
1357not necessarily, it will depend on the values of K and concentration of acids and their mixing proportion.
I don't have any example for that right now.
1@ eureka
plz dont go by above method of yours and it is far simpler to calculate for the mixing of weak acids which are fairly concentrated the formula you can use is √( K1C1+K2C2+K3C3...................)
here Ci ,C2 ...are the concentration after mixing
*remember the formula holds down for mixing of weak and fairly concentrated acids only
