What volume of 0.10 M sodium formate solution should be added to 50ml of 0.05 M formic acid to produce a buffer solution of pH 4.0? [ for formic acid is 3.80] A) 3.96 ml B) 25 ml C) 39.6 ml D) 100 ml
Answer
Verified
Hint: Buffer solution is basically an aqueous solution which resists the change in pH when a small amount of strong acid or base is added and formed by the mixture of weak acid and its conjugate base or its salt and vice-versa. The above question can be solved by the Henderson-Hasselbach equation.
Formula Used: For acidic buffer-$pH=p{{K}_{a}}+\log \dfrac{\text{Conjugate Base}}{\text{Weak Acid}}$
Complete answer: First, let us consider the volume of 0.10 M sodium formate added to be x. Now, with the help of mole concept, we have No. of moles in x ml of 0.10 M sodium formate = $\dfrac{0.10}{1000}$ $\times$ x= 0.0001x No. of moles in 50 ml of 0.05 M formic acid = $\dfrac{0.05}{1000}$ $\times$ 50 = 0.0025 In this M represents the molarity of solution. Thus, $\dfrac{sodium\;formate}{formic\;acid}$ = $\dfrac{0.0001x}{0.0025}$ = 0.04 x, represents the $\dfrac{salt}{acid}$ ratio. Now, from the Henderson’s equation, pH = pK$_a$+ log $\dfrac{salt}{acid}$ Given, pH = 4.0, pK$_a$= 3.80; Put these values in the Henderson’s equation 4.0 = 3.80 + log 0.04 x log 0.04 x = 0.2 Therefore, x = 39.6 ml
So, we can conclude that the volume of 0.10 M sodium formate added is 39.6 ml. The correct option is (C).
Note: Don’t get confused while solving the log. Here the logarithmic term is with the base 10. The pK$_a$ and pH terms are related to each other. The pH value at which the chemical species accept, or donate a proton is considered to be pK$_a$ value. The pK$_a$ value shows inverse relation with the acid, i.e. lower the pK$_a$, the stronger will be the acid.
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