1) 2.3 x 10 what is the answer

Chemistry

1 answer:

BlackZzzverrR [31]2 years ago

5 0

Answer:

2.3 x 10 =23

Explanation:

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Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.5

e-lub [12.9K]

Answer:

Buffer B has the highest buffer capacity.

Buffer C has the lowest buffer capacity.

Explanation:

An effective weak acid-conjugate base buffer should have pH equal to $pK_{a}$ of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.

Acetic acid is a weak acid and $CH_{3}COO^{-}$ is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):

$pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

$pK_{a}(CH_{3}COOH)=4.74$

Buffer A: $pH=4.74+log(\frac{0.10}{0.10})=4.74$

Buffer B: $pH=4.74+log(\frac{0.30}{0.30})=4.74$

Buffer C: $pH=4.74+log(\frac{0.10}{0.50})=4.04$

So, both buffer A and buffer B has same pH value which is also equal to $pK_{a}$ . Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.

The pH of buffer C is far away from $pK_{a}$ . Therefore, buffer C has the lowest buffer capacity.

6 0

3 years ago

Using the formula M1V1 = M2V2 , how many milliliters of a 2. 50 M hydrochloric acid solution is required to make 100. 0 mL of a

gladu [14]

Dilution of the solution can be calculated by the formula of the molarity and volume. The initial volume of 2.50 M solution was 30 mL.

<h3>What is the relationship between molar concentration and dilution?</h3>

Molar concentration or the dilution factor is in an inverse relationship and with an increase in the dilution, the molarity of the solution decreases.

Given,

Initial molarity = 2.50 M

initial volume = ?

Final molarity = 0.750 M

Final volume = 100.0 ml

Substituting values in the formula:

$\begin{aligned}\rm V_{1} &= \rm \dfrac{M_{2}V_{2}}{M_{1}}\\\\&= \dfrac{0.750 \times 100}{2.50}\\\\&= 30 \;\rm mL\end{aligned}$