1.ka for HF is 6.8x10^-4. calculate the kb for its conjugate base, the flouride

ID: 953551 • Letter: 1

Question

1.ka for HF is 6.8x10^-4. calculate the kb for its conjugate base, the flouride ion, F-

2. which of the following salts dissovled in water will form basic solutions?

NH4Cl? Cu(No3)2? NaCN? LiF?

3. what is the concentration (in M) of hydronium ions [H3O+] or [H+] in a solution at 25°C with pH=4.282

4. what is the pOH of a solution with a pH of 3.7?

5. the kb of methylamine is 4.4 x 10^-4. calculate tje pH of a 0.35 M (aq) solution of CH3NH2

6. the ka of (HClO) is 3.10 X 10^-8. what is the pH of a 0.0385 M solution of HClO

1.ka for HF is 6.8x10^-4. calculate the kb for its conjugate base, the flouride ion, F-

Kw = Ka*Kb

Kb = (Kw/Kb) = (10^-14)/(6.8*10^-4) = 1.470588*10^-11

NH4Cl --> Acidic

Cu(No3)2 --> Acidic

NaCN --> basic

LiF --> basic

pH = -logH+

[H+] = 10^-pH = 10^-4.282 = 0.00005223961

pOH = 14-ph = 14-3.7 = 10.3

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