A 500 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 5 mL of 1.00 M KOH. What is the pH following this addition

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lions [2.5K] · Answer 1 · 2026-02-21T08:21:13+03:00

Answer: The solution's pH post KOH addition is 3.84

Explanation:

Given information:

pH of buffer = 3.75

$pK_a$ of formic acid = 3.75

Employing the Henderson-Hasselbalch equation for formate buffer:

$pH=pK_a+\log(\frac{[HCOO-]}{[HCOOH]})$

Substituting values into the above equation yields:

$3.75=3.75+\log(\frac{[HCOO-]}{[HCOOH]})\\\\\log(\frac{[HCOO-]}{[HCOOH]})=0\\\\\frac{[HCOO-]}{[HCOOH]}=1$

$[HCOO-]=[HCOOH]$

Additional given information:

Concentration of formate buffer = 0.100 M

$[HCOO-]+[HCOOH]=0.1$

$[HCOO-]=[HCOOH]=0.05M$

As the buffer volume remains constant, the concentration reflects the moles of formate ions and formic acid.

To find the number of moles based on the specified molarity, utilize the following equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of KOH = 1.00 M

Volume of solution = 5 mL

Incorporating these values into the equation gives:

$1.00M=\frac{\text{Moles of KOH}\times 1000}{5mL}\\\\\text{Moles of KOH}=0.005mol$

The reaction between formic acid and KOH can be represented as:

$HCOOH+KOH\rightarrow HCOO^-+H_2O$

Initial: 0.05 0.005 0.05

Final: 0.045 - 0.055

The total volume of solution = 500 + 5 = 505 mL = 0.505 L (Note: 1 L = 1000 mL)

To determine the pH of the acidic buffer, we can use the provided equation from the Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[HCOO^-]}{[HCOOH]})$

Given information:

$pK_a$ = the negative logarithm of formic acid's dissociation constant = 3.75

$[HCOO^-]=\frac{0.055}{0.505}$

$[HCOOH]=\frac{0.045}{0.505}$

pH =?

Substituting values into the equation results in:

$pH=3.75+\log(\frac{0.055/0.505}{0.045/0.505})\\\\pH=3.84$

Consequently, the pH of the solution following KOH addition is 3.84