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elixir [45]
3 years ago
10

The ka values for several weak acids are given below. which acid (and its conjugate base) would be the best buffer at ph = 8.0?

Chemistry
2 answers:
Pie3 years ago
7 0
One of the best buffer choice for pH = 8.0 is Tris with Ka value of  6.3 x 10^-9.

To support this answer, we first calculate for the pKa value as the negative logarithm of the Ka value: 
     pKa = -log Ka

For Tris, which is an abbreviation for 2-Amino-2-hydroxymethyl-propane-1,3 -diol and has a Ka value of 6.3 x 10^-9, the pKa is
     pKa = -log Ka
            = -log (6.3x10^-9)
            = 8.2

We know that buffers work best when pH is equal to pKa:
     pKa = 8.2 = pH 

Therefore Tris would be a best buffer at pH = 8.0.
Alekssandra [29.7K]3 years ago
4 0

Tris and its conjugate base would be the best buffer at pH 8.

Further Explanation:

Buffer solution:

The solutions that oppose any change in their pH on addition of small amounts of acid or base are called buffer solutions. These are formed wither by a weak base and its conjugate acid or a weak acid and its conjugate base.

Henderson-Hasselbalch equation for a buffer solution is as follows:

{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{\text{Base}}} \right]}}{{\left[ {{\text{Acid}}} \right]}}     …… (1)                                                        

A buffer is most effective when the concentration of both acid and its conjugate base is the same. So equation (1) then modifies as follows:

{\text{pH}} = {\text{p}}{K_{\text{a}}}   …… (2)                                                                            

Therefore a buffer works best if its pH becomes equal to {\text{p}}{K_{\text{a}}}.

The formula to calculate {\text{p}}{K_{\text{a}}} is as follows:

{\text{p}}{K_{\text{a}}} =  - \log {K_{\text{a}}}  …… (3)                                                            

Here, {K_{\text{a}}} is the dissociation constant of acid.

Substitute \begin{aligned}{\text{p}}{K_{\text{a}}} &=- \log \left( {6.3 \times {{10}^{ - 9}}} \right)a\\&= 8.20\\\end{aligned} for {K_{\text{a}}} in equation (3) to calculate {\text{p}}{K_{\text{a}}} of tris.

 \begin{aligned}{\text{p}}{K_{\text{a}}} &=  - \log \left( {6.3 \times {{10}^{ - 9}}} \right)a\\&= 8.20\\\end{aligned}

Substitute 7.9 \times {10^{ - 7}} for {K_{\text{a}}} in equation (3) to calculate {\text{p}}{K_{\text{a}}} of MES.

 \begin{aligned}{\text{p}}{K_{\text{a}}} &=  - \log \left( {7.9 \times {{10}^{ - 7}}} \right)\\&= 6.10\\\end{aligned}

Substitute 1.8 \times {10^{ - 5}} for {K_{\text{a}}} in equation (3) to calculate {\text{p}}{K_{\text{a}}} of acetic acid.

\begin{aligned}{\text{p}}{K_{\text{a}}} &= - \log \left( {1.8 \times {{10}^{ - 5}}} \right)\\&= 4.74\\\end{aligned}

Substitute 1.8 \times {10^{ - 4}} for {K_{\text{a}}} in equation (3) to calculate {\text{p}}{K_{\text{a}}} of formic acid.

 \begin{aligned}{\text{p}}{K_{\text{a}}} &= - \log \left( {1.8 \times {{10}^{ - 4}}} \right)\\&= 3.74\\\end{aligned}

Substitute 3.2 \times {10^{ - 8}} for {K_{\text{a}}} in equation (3) to calculate {\text{p}}{K_{\text{a}}} of HEPES.

 \begin{aligned}{\text{p}}{K_{\text{a}}}&= - \log \left( {3.2 \times {{10}^{ - 8}}} \right) \\&= 7.49\\\end{gathered}

The value of {\text{p}}{K_{\text{a}}} of tris is nearly equal to the required pH. So tris along with its conjugate base would be the best buffer at pH 8.

Learn more:

Write the chemical equation responsible for pH of buffer containing   and  : brainly.com/question/8851686

Reason for the acidic and basic nature of amino acid. brainly.com/question/5050077

Answer details:

Grade: High School

Chapter: Acid, base and salts

Subject: Chemistry

Keywords: tris, acetic acid, formic acid, MES, HEPES, pH, pKa, Henderson, best buffer, 7.49, 8.20, 4.74, 3.74, Ka.

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7 0
3 years ago
If equal volumes of 0.1 M HCl and 0.2 M TRIS (base form) are mixed together. The pKa of TRIS is 8.30. Which of the following sta
blondinia [14]

Answer:

option D is correct

D. This solution is a good buffer.

Explanation:

TRIS (HOCH_{2})_{3}CNH_{2}

if TRIS is react with HCL it will form salt

(HOCH_{2})_{3}CNH_{2} + HCL ⇆   (HOCH_{2})_{3}NH_{3}CL

Let the reference volume is 100

Mole of TRIS is =  100 × 0.2 = 20

Mole of HCL is =  100 × 0.1 = 10

In the reaction all of the HCL will Consumed,10 moles of the salt will form

and 10 mole of TRIS will left

hence , Final product will be salt +TRIS(9 base)

H = Pk_{a} + log (base/ acid)

8.3 + log(10/10)

8.3

6 0
3 years ago
If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount
Schach [20]

<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL

<u>Explanation:</u>

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = 0.16 M

Given mass of manganese (II) nitrate tetrahydrate = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Putting values in above equation, we get:

0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL

Volume of water = Volume of solution = 398.4 mL

Hence, the amount of water required to prepare given amount of salt is 398.4 mL

4 0
3 years ago
An antacid tablet contains 640.0 mg of magnesium oxide per tablet.
Salsk061 [2.6K]

Answer:

317.6 mL

Explanation:

Step 1: Write the balanced neutralization equation

MgO + 2 HCl ⇒ MgCl₂ + H₂O

Step 2: Calculate the mass corresponding to 640.0 mg of MgO

The molar mass of MgO is 40.30 g/mol. The moles corresponding to 640.0 mg (0.6400 g) of MgO are:

0.6400 g × (1 mol/40.30 g) = 0.01588 mol

Step 3: Calculate the moles of HCl that react with 0.01588 moles of MgO

The molar ratio of MgO to HCl is 1:2. The moles of HCl are 2/1 × 0.01588 mol = 0.03176 mol

Step 4: Calculate the volume of 0.1000 M HCl that contains 0.03176 moles

0.03176 mol × (1 L/0.1000 mol) = 0.3176 L = 317.6 mL

3 0
3 years ago
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