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emmainna [20.7K]
3 years ago
9

Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

= 7.1 x 10-4) fH 0+1= x 10 M (Enter your answer in scientific notation.) pH =
Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer : The H_3O^+ ion concentration is, 1.12\times 10^{-3}M and the pH of a buffer is, 2.95

Explanation : Given,

K_a=7.1\times 10^{-4}

Concentration of HNO_2 (weak acid)= 0.26 M

Concentration of KNO_2 (conjugate base or salt)= 0.89 M

First 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 (7.1\times 10^{-4})

pK_a=4-\log (7.1)

pK_a=3.15

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

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

pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}

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

pH=3.15+\log (\frac{0.89}{0.26})

pH=2.95

The pH of a buffer is, 2.95

Now we have to calculate the H_3O^+ ion concentration.

pH=-\log [H_3O^+]

2.95=-\log [H_3O^+]

[H_3O^+]=1.12\times 10^{-3}M

The H_3O^+ ion concentration is, 1.12\times 10^{-3}M

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Answer:

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Explanation:

<u>Step 1: Define</u>

[Given] 57.5 L Xe at STP

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<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 57.5 \ L \ Xe(\frac{1 \ mol \ Xe}{22.4 \ L \ Xe})(\frac{6.022 \cdot 10^{23} \ atoms \ Xe}{1 \ mol \ Xe})
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How many moles of \rm CaCl_2 will be produced?

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<h3>What is density?</h3>

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