Calculate the pH of a 0.085 M solution of aniline, C_6H_5NH_2. What kind of prob

Question

Calculate the pH of a 0.085 M solution of aniline, C_6H_5NH_2. What kind of problem do you think this is? Calculate the pH of a 0.0015 M solution of sodium formate, HCOONa. What kind of problem do you think this is? Calculate the pH of a 0.020 M solution of ammonium chloride, NH4CI. What kind of problem do you think this is? Calculate the pH of a solution prepared by mixing 100.0 mL of 0.0100 M acetic acid (CH_3COOH) with 175.0 mL of 0.0100 M sodium acetate, CH_3COONa. What kind of problem do you think this is? A solution is prepared by mixing 100.0 mL of 0.100 M acetic acid (CH_3COOH) with 175.0 mL of 0.100 M sodium acetate, CH_3COONa. Determine the pH after the addition of 40.0 mL of 0.01 M HC1. What kind of problem do you think this is? A solution is prepared by mixing 100.0 mL of 0.100 M acetic acid (CH_3COOH) with 175.0 mL of 0.100 M sodium acetate, CH_3COONa. Determine the pH after the addition of 40.0 mL of 0.01 M NaOH. What kind of problem do you think this is? 11) Determine the volume of 0.182 M phosphoric acid, H_3PO_4, needed to titrate 239.86 mL of 0.197 M sodium hydroxide, NaOH, to the equivalence point? What kind of problem do you think this is?

Explanation / Answer

5)
kb of aniline = 4.3 * 10^-10
CH3NH2 + H2O ---> CH3NH3+ + OH-
0.085 0 0 (initial)
0.085-x x x (at equilibrium)

Kb = [CH3NH3+][OH-]/[CH3NH2]
4.3*10^-10 = x*x/(0.085-x)

since kb is small, x will be small and it can be ignored as compared to 0.085
Above expression now becomes,
4.3*10^-10 = x*x/(0.085)
x = 6.05*10^-6 M

so,
[OH-] = 6.05*10^-6 M

pOH = -log [OH-]
= -log (6.05*10^-6 M)
= 5.22

pH = 14 - pOH
= 14 - 5.22
= 8.78
Answer: 8.78

This is dissociation of weak acid

Only 1 question at a time please

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.