can help show how they got answers Consider the titration of 100. 0 mL of 0. 250

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can help show how they got answers

Consider the titration of 100. 0 mL of 0. 250 M aniline (Kb = 3. 82 times 10-10) with 0. 500 M HC1. Calculate the pH of the solution at the stoichiometric point. Calculate the pH at the equivalence point for the titration of 1. 0 M ethylamine, C2H5NH2, by 1. 0 M perchloric acid, HC1o4. (pKb for C2H5NH2 = 3. 25) A 100. 0-mL sample of 0. 2 M (CH ) N (kB = 5. 33 times 10-5) is titrated with 0. 2 MHC1. What is the pH at the equivalence point? Which of the following is the net ionic equation for the reaction that occurs during the titration of nitric acid with potassium hydroxide? Suppose a buffer solution is made from formic acid, HCHo2, and sodium formate, NaCHo2. What is the net ionic equation for the reaction that occurs when a small amount of sodium hydroxide is added to the buffer?

Explanation / Answer

a)

at eq point => n(A^-) = n(H^+) = 25 mmoles => V(HCl) = 50 mL,
where V(tot) = 150 mL

and at eq point we have a weak acid HA, there: [HA] = 25/150 M = 0,167 M

pH = 0.5*(pKa - lg(0.167)) = 2.68

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B)

[C2H5NH3+] = 1.0 M
Kb = 0.000562

C2H5NH3+ + H2O <---> C2H5NH2 + H3O+

K = Kw/Kb = 1.78 x 10^-11 = x^2 / 1-x

x = 4.22 x 10^-6

pH = 5.53

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C)

(CH3)3N + H+ >> (CH3)3NH+

Moles (CH3)3N = 0.100 L x 0.2 =0.02
Moles HCl needed = 0.02
volume HCl = 0.02 / 0.2 M =0.100 L

total volume = 0.200 L
concentration (CH3)3NH+ = 0.02 / 0.200 L =0.100 M

(CH3)3NH+ + H2O <----> (CH3)3N + H3O+

K = Kw/ Kb = 1.9 x 10^-10

1.9 x 10^-10 = x^2 / 0.1-x

x = 4.3 x 10^-6 M

pH = - log 4.3 x 10^-6 = 5.4

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D)

HI(aq)+KOH(aq)?H2O(l)+KI(aq)-

H+(aq) + I-(aq) + K+(aq) + OH-(aq) --> H2O(l) + K+(aq) + I-(aq)

H+(aq) + I-(aq) + K+(aq) + OH-(aq) --> H2O(l) + K+(aq) + I-(aq)

H+(aq) + OH-(aq) --> H2O(l) (Final Answer)

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E)

The buffer solution contains the following components:
HCHO2 molecules
H+ ions
CHO2- ions
Na+ ions

Adding NaOH effectively adds OH- ions to the mix. The Na+ ions are spectator ions and do not enter into the net ionic equation
The OH- ions will reacte with H+ ions, forming H2O, thereby reducing the concentration of H+ ions. The OH- ions will also react with HCHO2 forming H2O and CHO2- ions. Since the Ka for the dissociation of HCHO2 is small, there is much more HCHO2 than H+, so the second reaction dominates. The net ionic reaction is then

HCHO2(aq) + OH-(aq) ---> H2O(l) + CHO2-(aq)

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