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3 years ago

Consider the following reaction:

Chemistry

1 answer:

Goshia [24]3 years ago

3 0

Answer:

2.01

Explanation:

First, let's convert grams to moles

(Na) 22.99 + (F) 18.998 = 41.988

Every mole of NaF is 41.988 grams

21/41.988 = 0.500143 moles of NaF

For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF

0.500143/2 = 0.250071

molarity = moles/liters

0.250071/0.125 = 2.0057 M

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Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE

nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let " $\alpha$ is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

$HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}$

Initial C 0 0

Equilibrium c(1- $\alpha$ ) c $\alpha$ c $\alpha$

Dissociation constant = $Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}$

$=\frac{c\alpha^{2}}{(1-\alpha)}$

In this case weak acids $\alpha$ is very small so, (1- $\alpha$ ) is taken as 1.

$Ka=C\alpha^{2}$

$\alpha=\sqrt\frac{ka}{c}$

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Substitute the given value.

$\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}$

$[H_{3}O^{+}]=c\alpha$

$[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}$

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3 years ago

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