If 50.0 g of formic acid (HCHO2, ka = 1.8 x 10^-4) and 30.0 g of sodium formate (NaCHO2) are dissolved to make 500 ml. of solution, the ph of this solution is 3.35.
Therefore, option C is the correct option.
Given,
Given mass of sodium formate = 30 g
Given mass of formic acid = 50 g
Volume of sodium formate = 500 ml
Volume of formic acid = 500ml
Molar mass of sodium formate = 68 g
Molar mass of formic acid = 46 g
<h3>To calculate concentration of sodium formate and formic acid</h3>
The ratio of number of moles and the volume of solution is Molar concentration of substance.
Concentration of sodium formate
Cb = 30/(68×500)
= 0.00088m
Concentration of formic acid
Ca = 50/(46×500)
= 0.00217m
Now,
by using Henderson hesselbalch equation,
pH = pKa + log(Cb/Ca)
pKa = -log(1.8 × 10^(-4))
= 3.75
pH = 3.75 + log(0.00088/0.00217)
pH = 3.75 - 0.392
pH = 3.35
Thus, we calculated that the value of pH of solution of formic acid and sodium formate is 3.35.
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(Lindlar catalyst) and
, alkynes are reduced to alkene.
