Respuesta :
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.
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
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