Determine the pH for the following salts in water. Assume that you have a 0.20 M

Question

Determine the pH for the following salts in water. Assume that you have a 0.20 M solutions of each of the salts. a. NaCN b. pyridinium bromide c. Li_2CO_3 For letters d, e, f, and g find the pH of the resulting solution assuming that you have 0.20 M of the stated salt and you also have 0.10 M of its conjugate. You will determine what the conjugate species should be (see the example in (d)). d. CH_3NH_3Cl (i.e. you have the acid, CH_3NH_3^- 0.20 M, so you need the conjugate base CH_3NH_3, 0.10 M) e. KH_2PO_4 (use as an acid, and stale its conjugate base) f. Li_2HAsO_4 (use as a base, and state its conjugate acid) g. NH_3 and its conjugate acid

Explanation / Answer

Q4.

pH for salt in aM = 0.2

a)

NaCN -> Na+ + CN-

CN- + H2O <-> HCN + OH-

Kb = [HCN][OH-]/[CN-]

Kb = Kw/Ka = (10^-14)/(4.9*10^-10) = 2.0408*10^-5

so

2.0408*10^-5 = x*x/(0.2-x)

x = OH = 0.002

pOH = -log(0.002)= 2.6989

pH = 14-2.6989 = 11.3011

b)

KB = 1.77*10^-9

so

Pyridinum Bromide --> Pyridinum ion + Bromide ion

Pyridinum ion + H2O <--> Pyridine + H3O+

Ka = [Pyridine ][H3O+] / [Pyridinum ion]

Ka = Kw/Kb = (10^-14)/(1.77*10^-9) = 6.497*10^-6

Ka = x*x/(M-x)

6.497*10^-6 = x*x/(0.2-x)

x = H3O+ = 0.001136

pH = -log(0.001136) = 2.944

c)

finally

for

Li2CO3:

2Li+ + CO3-2

CO3-2 + H2O <-> HCO3- + OH-

KB = [HCO3-][OH-]/[CO3-2]

4.7 x 10-11 = x*x/(0.2-x)

x = OH- = 3.067*10^-6

pOH = -log( 3.067*10^-6) = 5.513

pH = 14-5.513 = 8.487

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