One litre of a solution with pH = 3 is mixed with nine litres of a solution with pH = 7.
(i) What is the pH of the resulting solution ?
(ii) What will be the pH of the solution if further 90 litres of water are added to the above solution,
(iii) What will be the pH of the solution if instead of water 10 litres of a solution with pH = 4 are added to
solution in (i)?
Answers
Answer:
Explanation:We will assume that mixing the two solutions mentioned in the question does not result in any chemical reaction and we will only consider the effects of dilution.
The answer to the question then depends on whether the solution of pH = 3 is a solution of a (I) strong acid, (ii) a weak acid or (iii) a buffer solution. We will assume that the solution of pH =7 is pure water, or a dilute solution of a salt of a strong acid and a strong base,e.g., NaCl.
(I) solution of a strong acid
Mixing 1L of solution of a strong acid with 9 L of pure water or a dilute solution of a salt ( say NaCl) reduces the concentration of H+ ion to one-tenth so the pH increases by one unit from 3 to 4.
(ii) solution of a weak acid
Ten times dilution of a weak acid solution does not reduce the H+ ion concentration to one-tenth of the original. If the initial concentration of the weak acid is C1 mol/L, and K is its dissociation constant . then [H+]1= (K*C1)^1/2
If C2 is the concentration of the diluted solution , [H+]2=(K*C2)^1/2
[H+]2/[H+]1 ={(K*C2)^1/2}/{(K*C1)^1/2} =(C2/C1)^1/2
In the present question C2 = C1/10
[H+]2/[H+]1 = 1/10^1/2 = 0.32
[H+]2 = 0.32*[H+]1
(pH)2 = (pH)1 - log 0.32 = (pH)1 + 0.49
Now (pH)1 = 3, so (pH)2 =3.49
Thus in the case of a weak acid the pH increases by about 0.5 . This answer assumes that in spite of ten times dilution the degree of dissociationof the weak acid is negligible in comparison with 1.
(iii) a buffer solution
The pH of a buffer solution does not change on dilution, so the pH remains 3.
The effects on pH of Adding 90 L of water after the first dilution can be easily calculated in the light of the above.
Answer:The answer to the question then depends on whether the solution of pH = 3 is a solution of a (I) strong acid, (ii) a weak acid or (iii) a buffer solution. We will assume that the solution of pH =7 is pure water, or a dilute solution of a salt of a strong acid and a strong base,e.g., NaCl.
(I) solution of a strong acid
Mixing 1L of solution of a strong acid with 9 L of pure water or a dilute solution of a salt ( say NaCl) reduces the concentration of H+ ion to one-tenth so the pH increases by one unit from 3 to 4.
ii) solution of a weak acid
Ten times dilution of a weak acid solution does not reduce the H+ ion concentration to one-tenth of the original. If the initial concentration of the weak acid is C1 mol/L, and K is its dissociation constant . then [H+]1= (K*C1)^1/2
If C2 is the concentration of the diluted solution , [H+]2=(K*C2)^1/2
[H+]2/[H+]1 ={(K*C2)^1/2}/{(K*C1)^1/2} =(C2/C1)^1/2
In the present question C2 = C1/10
[H+]2/[H+]1 = 1/10^1/2 = 0.32
[H+]2 = 0.32*[H+]1
(pH)2 = (pH)1 - log 0.32 = (pH)1 + 0.49
Now (pH)1 = 3, so (pH)2 =3.49
Thus in the case of a weak acid the pH increases by about 0.5 . This answer assumes that in spite of ten times dilution the degree of dissociationof the weak acid is negligible in comparison with 1.
iii) a buffer solution
The pH of a buffer solution does not change on dilution, so the pH remains 3.
The effects on pH of Adding 90 L of water after the first dilution can be easily calculated in the light of the above