pH Before =pH After = Calculate the pH of 100.0 mL of a buffer that is 0.0700 M NH4CI and 0.110M NH3 before andafter the addition of 1.00 mL of 5.90 M HNO3.

Answered: Calculate the pH of 100.0 mL of a…

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter17: Principles Of Chemical Reactivity: Other Aspects Of Aqueous Equilibria

Section17.2: Controlling Ph: Buffer Solutions

Problem 17.5CYU: Calculate the pH of 0.500 L of a buffer solution composed of 0.50 M formic acid (HCO2H) and 0.70 M...

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pH Before =

pH After =

Calculate the pH of 100.0 mL of a buffer that is 0.0700 M NH4CI and 0.110M NH3 before and
after the addition of 1.00 mL of 5.90 M HNO3.

Expert Solution

Step 1

Henderson Hasselbalch equation

pH of a buffer is calculated using Henderson Hasselbalch equation.

For a basic buffer,

$pOH = {pK}_{b} + \log \frac{[salt]}{[base]} [salt] = concentration of salt [base] = concentration of base$

Step 2

pH before addition

$\begin{array}{rcl} pOH & = & {pK}_{b} + \log \frac{[salt]}{[base]} \\ K_{b} & = & 1.78 \times 10^{- 5} \\ pKb & = & - \log (K_{b}) \\ = & - \log (1.78 \times 10^{- 5}) \\ = & 4.74 \\ [salt] & = & 0.07 M \\ [base] & = & 0.11 M \\ On substitution, \\ pOH & = & {pK}_{b} + \log \frac{[salt]}{[base]} \\ = & 4.74 + \log \frac{0.07}{0.110} \\ = & 4.54 \\ pH + pOH & = & 14 \\ pH & = & 14 - pOH \\ = & 14 - 4.54 \\ pH & = & 9.46 \end{array}$

Step 3

On adding 1mL 5.9 M HNO₃,

$concentration of {HNO}_{3} = 5.90 M$

On adding the acid, the ammonia reacts with the H⁺ ions.

${NH}_{3} + H_{3} O^{+} \to {NH}_{4} OH + H_{2} O$

The concentration in terms of number of moles are;

$\begin{array}{rcl} number of moles of {NH}_{3} & = & molarity \times total volume \\ = & 0.07 \times 100 \\ = & 7 \end{array}$

${NH}_{3} + H_{3} O^{+} \to {NH}_{4}^{+} + H_{2} O I 7 5.9 10 C - 5.9 - 5.9 + 5.9 E 1.18 0 15.9$

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