3 years ago

Please help! (no links)

Chemistry

1 answer:

Rus_ich [418]3 years ago

5 0

It's D because I know bcz I'm smart 8 times 8 is 60-nart

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A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles

alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of $HF$ .

$\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}$

$\text{Moles of HF}=0.250M\times 1.50L=0.375mol$

Now we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

The reaction will be:

$HF+OH^-\rightleftharpoons F^-+H_2O$

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At eqm. (0.375-0.100) 0 (0.375+0.100)

= 0.275 = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[F^-]}{[HF]}$

Now put all the given values in this expression, we get:

$pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]$

$pH=3.41$

Thus, the pH of the solution is, 3.41

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