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1. A solution is prepared by adding 100 mL of 0.2 M hydrochloric acid to 100...

1. A solution is prepared by adding 100 mL of 0.2 M hydrochloric acid to 100 mL of 0.4 M sodium formate. Is this a buffer solution, and if so, what is its pH?

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Answer #1

Sodium formate --> HCOONa

HCOONa -----> HCOO- + Na+

HCOO- + HCl ----------> HCOOH

In another form we can write as

HCOOH --------> HCl + HCOO-

Moles of HCOO- = Molarity*V in L = 100*0.4/1000 = 40 millimoles

Moles of HCl = 0.2*100/1000 = 20 millimoles

HCOOH --------> HCl + HCOO-

0 20 40

20 0 20

pH = pKa + log [HCOONa] / [HCOOH]

Since acidic buffer is formed after adding HCl

we are using the formula

pH = pKa + log [salt] / [Acid]

pH = pKa + log [20 /20]

pH = pKa = 3.75 [since pKa of formic acid is 3.75]

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