To produce an ideal buffer with a pH = 5 plusminus 1, solution of and could be u

Question

To produce an ideal buffer with a pH = 5 plusminus 1, solution of and could be used. CH_2NH_2, CH_3 NH_3 I^- (K_b for CH_2NH_2 = 5.0 times 10^-4) NH_2NH_3^-Br^-, NH-2NH_2 (K_b for NH_2 NH_2 = 1.3 times 10^-5) HBr, NH_4Br (K_b for NH_3 = 1.8 times 1)^-5) HF, C_5F (K-b for HF = 7.2 times 10^-4) C_6H_5NH_2, C-6H_5NH_3^+Br^- (K-b for C_6H_5NH_3 = 4.2 times 10^10) Classify the following aqueous solution as acidic, basic, or natural. Given that K_b of C_6H_5NH_2 is 4.2 times 10^-16 and that K_b of HClO is 3.5 times 10^-5 CH_3NH_3^+BrO^-_4 RbNO_3 CsCN C_6H_5NH^+_5ClO^- Determine the pH of a 1.45 M solution of C_6H_5NH_3ClO^-_4 (K_b of C_6H_5NH_2 = 4.2 = 10^-10) In the titration of 40.0 mL of 0.15 M HBr with 0.20 M CsOH, determine the pH after the addition of 14.0 mL of CsOH.

Explanation / Answer

Q18

pH = 5+/-1

we need pKa = pH approx

which is ACIDIC

so ignore all bases

also, a buffer is a weak acid + conjugate base

the best solution:

HF and CsF, since pKa = 3.7 approx

Q19.

a)

CH3NH3 and BrO4-

CH3NH3+ in solution will:

CH3NH3+ + H2O <-> CH3NH2 + H3O+, which is ACIDIC

b)

RbNO3 --> neutral ions

c)

CsCN --> Cs+ and CN-

CN- + H2O <-> HCN ´+ OH-, which is BASIC

d)

the Kb > Ka, therefore, this is more acidic... but in reallity pH is near 7

Q20

Kb [H+][C6H5NH3+][OH-] /[ C6H5NH2]

Kb = (x*x)/(1.45-x)

4.2-*10^-1= = x*x / (1.45-x)

x = 2.46*10^-5

pOH = -log( 2.46*10^-5) = 4.60

ph = 14-pOH = 14-4.60 = 9.4

Q21.

mol of acid = MV = 40*0.15 = 6 mmol of acid

mmol of base = MV = 0.2*14 = 2.8 m

mmol of aic dleft = 6-2.8 = 3.2 mmol

VT = 15+14 = 29 ml

then...

[H+] = 3.2/29 = 0.110

pH = -log(0.110 = 0.9586

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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