An analytical chemist is titrating 132.3 mL of a 0.7100 M solution of methylamine (CH,NH2 with a 0.7500 M solution of HNO 3* The p K, of methylamine is 3.36. Calculate the pH of the base solution after the chemist has added 111.4 mL of the HNO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added Round your answer to 2 decimal places.

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29-10-2019
Chemistry

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An analytical chemist is titrating 132.3 mL of a 0.7100 M solution of methylamine (CH,NH2 with a 0.7500 M solution of HNO 3* The p K, of methylamine is 3.36. Calculate the pH of the base solution after the chemist has added 111.4 mL of the HNO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added Round your answer to 2 decimal places.

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Please answer the question.

OlaMacgregor OlaMacgregor · Answer 1 · 2019-10-29T05:55:54+01:00

Explanation:

Calculate the moles of [tex]CH_{3}NH_{2}[/tex] as follows.

= 0.1323 L x 0.7100 M

= 0.0940 mol

Now, calculate the moles of [tex]HNO_{3}[/tex] as follows.

= 0.1114 L x 0.7500 M

= 0.0836 mol

According to the ICF table,

[tex]CH_{3}NH_{2} + H^{+} \rightarrow CH_{3}NH_{3}^{+}[/tex]

Initial : 0.0940 mol 0.0836mol 0.0 mol

Change: -0.0836 -0.0836 +0.0836

Final: 0.0104 mol 0.0 mol 0.0836 mol

According to Henderson-Hasselbalch equation, we have

[tex]pOH = pK_{b} + log \frac{[CH_{3}NH_{3}^{+}]}{[CH_{3}NH_{2}]}[/tex]

Putting the given values into the above equation we get the value of pOH as follows.

[tex]pOH = pK_{b} + log \frac{[CH_{3}NH_{3}^{+}]}{[CH_{3}NH_{2}]}[/tex]

pOH = 3.36 + log [tex]\frac{0.0836}{0.0104}[/tex]

= 3.36 + 0.905

= 4.265

Therefore, calculate the pH as follows.

pH = 14 - 4.265

= 9.7

Thus, we can conclude that the pH of the given base solution is 9.7.

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-01T12:53:50+02:00

The pH of the base solution after the chemist has added 111.4 mL of the HNO3 solution is mathematically given as

pH= 9.7

What is the pH of the base solution after the chemist has added 111.4 mL of the HNO3 solution to it.?

Question Parameter(s):

An analytical chemist is titrating 132.3 mL of a 0.7100 M solution

The p K, of methylamine is 3.36.

Generally, the equation for the Chemical Reaction is mathematically given as

CH3NH2 + H^{+}----> CH_{3}NH3^{+}

Therefore

[tex]pOH = pK_{b} + log \frac{[CH_{3}NH_{3}^{+}]}{[CH_{3}NH_{2}]}[/tex]

[tex]pOH = 3.36 + log \frac{0.0836}{0.0104}[/tex]

pOH= 4.265

In conclusion, The pH is

pH = 14 - 4.265

pH= 9.7