Q. A buffer is prepared by mixing 0.65 moles of weak acid (HA) and 0.55 moles of
ID: 1036308 • Letter: Q
Question
Q. A buffer is prepared by mixing 0.65 moles of weak acid (HA) and 0.55 moles of corresponding conjugate base (NaA). Consider the Ka value of the weak acid as 2.95 x10-5. Find out the effect of adding the following amount of acids or bases to this buffer.Consider the original volume of buffer as 1L, and there is no change in volume by additionof acids or bases.
1. 0.1 mole of HNO3
2. 0.087 moles of H3PO4
3. 0.22 moles of Ba(OH)2
4. 0.33 moles of LiOH
Calculate the
1. pH of the original buffer,
2. What would the change in pH of the buffer when the acid/base mentioned aboveare added separately?
3. Calculate the pH change when these acid/base are added to water separately.
Explanation / Answer
1. pH of the original solution is
pH = pKa + log([salt]/[acid]) = pKa + log(nsalt/V/nacid/V)
Ka = 2.95 x10-5 Therefore pKa = -log ka = -log (2.95 x10-5) =4.53
A buffer is prepared by mixing 0.65 moles of weak acid (HA) and 0.55 moles of corresponding conjugate base (NaA)
Threfore pH = 4.53 + log (0.55/0.65) here volume is 1 L so it is cancelled out.
pH = 4.45
As per defination, pH value should not change when small quantity of acid or base is added in buffer solution. So one way of anwering the 1-4 is pH value will not change.
Second way of answering the question is given below:
1. 0.1 mole of HNO3 is added in original buffer solution:
Thus moles of acid and salts will change w.r.t. original buffer:
Acid: (0.55 mole + 0.1 mole) = 0.65 mole
Salt: (0.65 mole - 0.1 mole) = 0.55 mole
pH = 4.53 + log (0.65/0.55) = 4.60
2. 0.087 moles of H3PO4 is added in original buffer solution:
Thus moles of acid and salts will change w.r.t. original buffer:
Acid: (0.55 mole + 0.087 mole) = 0.637 mole
Salt: (0.65 mole - 0.087 mole) = 0.563 mole
pH = 4.53 + log (0.637/0.563) = 4.58
3. 0.22 moles of Ba(OH)2 is added in original buffer solution:
Thus moles of acid and salts will change w.r.t. original buffer:
Acid: (0.55 mole - 0.22 mole) = 0.33 mole
Salt: (0.65 mole + 0.22 mole) = 0.87 mole
pH = 4.53 + log (0.33/0.87) = 4.11
4. 0.33 moles of LiOH is added in original buffer solution:
Thus moles of acid and salts will change w.r.t. original buffer:
Acid: (0.55 mole - 0.33 mole) = 0.22 mole
Salt: (0.65 mole + 0.33 mole) = 0.98 mole
pH = 4.53 + log (0.22/0.98) = 3.88
2. Change in pH of original buffer solution after addition of acid increases whereas pH of original buffer solution after addition of base decreases. This indicates how buffer solution resists drastic change in the pH value.
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