please show steps on how to solve. 21 Calculate the pH od a solution formed by m

Question

please show steps on how to solve.

21 Calculate the pH od a solution formed by mixing 1000 mL of 0.20 M HOO with 2000 mL. of 0.30 M KCIO The Ka for 1CCO is 29- A) 801 B) 646 C7.54 DI 7.71 E) 5.99 3) A solution contairs 0021 MCI and 0017 M F. A solution containing opper () ions is added to selectively precipitate orne of the ions. At what concentration of copper (1) ion will a precipitate begin to form? What is the identity of the precipitate? Ksp/Cu) LD 10-e, k,pical) = 5 1x 10-12 A) 30 10-10 M. Cul B) 48 10-5 M, Cul C)30 10-10 M, CuCI D) 4.8 x 10-5 M, CuCl E) No precipitate will form at any concentration of copper ().

Explanation / Answer

2)

we know that

moles = molarity x volume (ml) / 1000

so

moles of HCl0 added = 0.2 x 100 / 1000 = 20 x 10-3

moles of KClO added = 0.3 x 200 / 1000 = 60 x 10-3

now

final volume = 100 + 200 = 300 ml

now

[HCl0] = 20 x 10-3 x 1000 / 300 = 2/30

[KCl0] = 60 x 10-3 x 1000 / 300 = 6/30

we know that

HClO is a weak acid and KClO is salt of its conjugate base

so

they form a buffer solutioon

now

for buffers

pH = pKa + log [KCl0 / HCl0]

also

pKa = -log Ka

so

pH = -log 2.9 x 10-8 + log [ (6/30) / (2/30)]

pH = 7.5376 + log (6/2)

pH = 8.0147

so

the answer is A) 8.01

3)

first consider CuCl

CuCl --> Cu+ + Cl-

the solubility product constant is given by

Ksp = [Cu+] [Cl-]

1 x 10-6 = [Cu+] [ 0.021]

[Cu+]= 4.762 x 10-5 M

now

consider CuI

CuI --> Cu+ + I-

the solubility product constant is given by

Ksp = [Cu+] [I-]

5.1 x 10-12 = [Cu+] [ 0.017]

[Cu+]= 3 x 10-10 M

so

precipitate of CuCl forms when [Cu+] = 4.762 x 10-5 M

precipitate of CuI starts to form when [Cu+] = 3 x 10-10 M

so

CuI is the first one to precipitate

so

the answer is

A) 3 x 10-10 M , CuI

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