Answer:
4·234 g
Explanation:
Given
Volume of the solution = 150 mL
Concentration of formic acid = 0·42 mol/L
pH of the buffer solution = 3·74
[tex]K_{a}[/tex] for HCOOH = 0·00018
∴ [tex]pK_{a}[/tex] of weak acid = -log([tex]K_{a}[/tex]) = -log(0·00018) = 3·745
Let the mass of solid sodium formate that must be added be m g
∴ The concentration of sodium formate = ((m ÷ 68·01) ÷ 150) × 10000
In this case conjugate base is sodium formate and weak acid is formic acid
3·74 = 3·745 + log(((m ÷ 68·01) ÷ 150) × 1000) - log(0·42)
⇒ -0·005 +log(0·42) = log(((m ÷ 68·01) ÷ 150) × 1000)
⇒ -0·005 - 0·377 = log(((m ÷ 68·01) ÷ 150) × 1000)
⇒ -0·382 = log(((m ÷ 68·01) ÷ 150) × 1000)
⇒ antilog(-0·382) = ((m ÷ 68·01) ÷ 150) × 1000
⇒ 0·415 × [tex]10^{-3}[/tex] = ((m ÷ 68·01) ÷ 150
⇒ 0·415 × [tex]10^{-3}[/tex] × 150 × 68·01 = m
∴ m = 4·234 g
∴ Mass of solid sodium formate that must be added is 4·234 g
A buffer solution formed by 150 mL of 0.42 mol/L formic acid, requires 4.3 g of solid sodium formate to have a pH of 3.74.
We have a buffer system formed by 150 mL of 0.42 mol/L formic acid (weak acid) and an unknown concentration of formate (conjugate base).
We can calculate the concentration of formate ion using the Henderson Hasselbach's equation.
[tex]pH = pKa + log \frac{[HCOO^{-} ]}{[HCOOH]} \\\\3.74 = -log(0.00018) + log \frac{[HCOO^{-} ]}{(0.42)} \\[HCOO^{-} ] = 0.42 M[/tex]
Sodium formate is a strong electrolyte, so its concentration must be the same as formate, i.e. 0.42 M.
We can calculate the required mass of sodium formate using the following expression.
[tex][HCOONa] = \frac{mass\ HCOONa}{MW(HCOONa) \times liters\ of\ solution} \\\\mass\ HCOONa = [HCOONa] \times MW(HCOONa) \times liters\ of\ solution\\\\mass\ HCOONa = \frac{0.42mol}{L} \times \frac{68.01g}{mol} \times 0.150 L = 4.3 g[/tex]
A buffer solution formed by 150 mL of 0.42 mol/L formic acid, requires 4.3 g of solid sodium formate to have a pH of 3.74.
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