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vovangra [49]
3 years ago
12

Determine the pH at the equivalence point for the titration of 40.0 mL of 0.1 M HNO2 by 0.200 M KOH (Example 16.7). (The pKa of

nitrous acid is 3.36.)
Chemistry
1 answer:
Fynjy0 [20]3 years ago
8 0

Answer:

pH=13.28

Explanation:

Nitrous acid

HNO_2 \longrightarrow H^+ + NO_2^-

Ka=\frac{[H^+][NO_2^-]}{[HNO_2]}

pKa=-log(Ka)=3.36

Ka=4.36*10^{-4}

<u>Balance: </u>

[H^+]=X

[NO_2^-]=X

[HNO_2]=0.1-X

Ka=\frac{X*X}{0.1-X}

4.36*10^{-4}=\frac{X^2}{0.1-X}

4.36*10^{-5}-4.36*10^{-4}*X=X^2

Solving:

X=6.4*10^{-3}

The NaOH is in excess:

[NaOH]_f=0.2-0.0064=0.1936

pH=14+log([OH^-])

pH=14+log(0.1936)=13.28

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The term concentration has to do with the amount of substance in solution. The concentration can be measured in several units. Generally, concentration is expressed in molarity, molality, mass concentration units or percentage.

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(a)  The given data is as follows.  

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Now, putting the given values into the above formula as follows.

                HB = \frac{2P}{\pi D [D - \sqrt{(D^{2} - d^{2})}]}                            = \frac{2 \times 1000 kg}{3.14 \times 10 mm [D - \sqrt{((10 mm)^{2} - (2.50)^{2})}]}  

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