Respuesta :
Given:
Buffer:
Concentration of acid = 0.5 m HCNO
Concentration of dissociated acid = 0.5 m CNO-
Volume of buffer = 100 ml
Volume of base = 100 ml
Concentration of NaOH (base) = 0.2 m
Ka(HCNO) = 3.5x10^-4
The formula for pH buffers:
pH = pKa - log([HA]/[A-])
ph = -log(3.5x10^-4) - log(0.5)
pH = 3.76
Buffer:
Concentration of acid = 0.5 m HCNO
Concentration of dissociated acid = 0.5 m CNO-
Volume of buffer = 100 ml
Volume of base = 100 ml
Concentration of NaOH (base) = 0.2 m
Ka(HCNO) = 3.5x10^-4
The formula for pH buffers:
pH = pKa - log([HA]/[A-])
ph = -log(3.5x10^-4) - log(0.5)
pH = 3.76
The pH of 100mL of buffer of 0.5 M HCNO and 0.5 M [tex]{\text{CN}}{{\text{O}}^ - }[/tex] on addition of 100 mL of 0.2 M NaOH solutionis [tex]\boxed{3.83}[/tex].
Further Explanation:
The aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid is termed as buffer solution. These solutions offer strong resistance to any change in their pH on addition of small quantity of strong acid or base.
Henderson-Hasselbalch equation:
This equation helps in determining the pH of buffer solution. Its mathematical form is given as follows:
[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}[/tex] …… (1)
Here,
[tex]\left[ {{{\text{A}}^ - }} \right][/tex] is concentration of conjugate base.
[HA] is concentration of acid.
Given mixture is a buffer solution of HCNO and [tex]{\text{CN}}{{\text{O}}^ - }[/tex]. Therefore Henderson-Hasselbalch equation becomes as follows:
[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{\text{CN}}{{\text{O}}^ - }} \right]}}{{\left[ {{\text{HCNO}}} \right]}}[/tex] …… (2)
The expression for [tex]{\text{p}}{K_{\text{a}}}[/tex] is as follows:
[tex]{\text{p}}{K_{\text{a}}} = - \log {K_{\text{a}}}[/tex] …… (3)
Where [tex]{K_{\text{a}}}[/tex] is the dissociation constant of acid.
Substitute [tex]3.5 \times {10^{ - 4}}[/tex] for [tex]{K_{\text{a}}}[/tex] in equation (3).
[tex]\begin{aligned}{\text{p}}{K_{\text{a}}} &=- \log \left( {3.5 \times {{10}^{ - 4}}} \right) \\&= 3.46 \\\end{aligned}[/tex]
Initial moles of [tex]{\text{CN}}{{\text{O}}^ - }[/tex] can be calculated as follows:
[tex]\begin{aligned}{\text{Moles of CN}}{{\text{O}}^ - } &= \left( {0.5{\text{ M}}} \right)\left( {{\text{100 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\&= 5{\text{ mol}} \\\end{aligned}[/tex]
Initial moles of HCNO can be calculated as follows:
[tex]\begin{aligned} {\text{Moles of HCNO}} &= \left( {0.5{\text{ M}}} \right)\left( {{\text{100 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\ &= 5{\text{ mol}} \\\end{aligned}[/tex]
Moles of NaOH can be calculated as follows:
[tex]\begin{aligned}{\text{Moles of NaOH}} &= \left( {0.2{\text{ M}}} \right)\left( {{\text{100 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right) \\ &=2{\text{ mol}} \\\end{aligned}[/tex]
When 2 moles of NaOH are added to the buffer solution, 2 moles of HCNO is neutralized while the same amount of [tex]{\text{CN}}{{\text{O}}^ - }[/tex] is formed. Since volumes are additive, total volume becomes 200 mL (100 mL + 100 mL).
Therefore concentration of [tex]{\text{CN}}{{\text{O}}^ - }[/tex] can be calculated as follows:
[tex]\begin{aligned}\left[ {{\text{CN}}{{\text{O}}^ - }} \right] &= \frac{{\left( {5 + 2} \right){\text{ mol}}}}{{\left( {200{\text{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)}} \\ &= 35{\text{ M}} \\ \end{aligned}[/tex]
Therefore concentration of HCNO can be calculated as follows:
[tex]\begin{aligned}\left[ {{\text{HCNO}}} \right] &= \frac{{\left( {5 - 2} \right){\text{ mol}}}}{{\left( {{\text{200 mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\text{ L}}}}{{1{\text{ mL}}}}} \right)}} \\ &= 15{\text{ M}} \\\end{aligned}[/tex]
Substitute 15 M for [HCNO], 35 M for [tex]\left[ {{\text{CN}}{{\text{O}}^ - }} \right][/tex] and 3.46 for [tex]{\text{p}}{K_{\text{a}}}[/tex] in equation (2).
[tex]\begin{aligned}{\text{pH}} &= 3.46 + {\text{log}}\left( {\frac{{35{\text{ M}}}}{{15{\text{ M}}}}} \right) \\&= 3.83 \\\end{aligned}[/tex]
Learn more:
- The reason for the acidity of water https://brainly.com/question/1550328
- Reason for the acidic and basic nature of amino acid. https://brainly.com/question/5050077
Answer details:
Grade:High School
Subject: Chemistry
Chapter: Acid, base and salts
Keywords: pH, buffer, pKa, 3.83, 3.46, HCNO, CNO-, 35 M, 15 M, 200 mL, 100 mL, 2 mol, 5 mol.
