1.- What is the pH of a solution prepared by mixing 100.00 mL of 0.020 M Ca(OH)2

Question

1.- What is the pH of a solution prepared by mixing 100.00 mL of 0.020 M Ca(OH)2 with 50.00 mL of 0.200 M NaOH? Assume that the volumes are additive.

A) 12.97 B) 13.45 C) 12.90 D) 13.15

I`m getting 12.74 and I do not know what I`m doing wrong.

2.- Kc is 1.67*10^20 at 25 C for the formation of iron(III) oxalate complex ion:

if .02 M Fe^3+ is initially mixted with 1.00 M oxalate ion, what is the concentration of Fe^3+ ion at equilibrium?

(A) 1.44*10^-22 M

(B) 8.35*10^19 M

(C) .01 M

(D) 6.94*10^21 M

Explanation / Answer

Q1) What is the pH of a solution prepared by mixing 100.00 mL of 0.020 M Ca(OH)2 with 50.00 mL of 0.200 M NaOH? Assume that the volumes are additive.

Solution :-

Ca(OH)2 and NaOH both are strong base

Lets first calculate moles of the each using their molarity and volume

Moles = molarity * volume in liter

Moles of Ca(OH)2 =0.020 mol per L * 0.100 L = 0.002 mol

1 mol Ca(OH)2 gives 2 mol OH- therefore moles of OH-

=0.002 mol Ca(OH)2 * 2 mol OH- /1 mol Ca(OH)2 = 0.004 mol OH-

Now lets calculate moles of NaOH

Moles of NaOH = 0.200 mol per L * 0.050 L =0.01 mol NaOH

1 mol NaOH gives 1 mole OH- therefore moles of OH- obtained from NaOH = 0.01 mol OH-

Now lets calculate total moles of OH- = 0.004 mol + 0.01 mol =0.014 mol OH-

Now we need to calculate new molarity of OH- at total volume

Total volume = 100 ml + 50 ml = 150 ml = 0.150 L

New molarity of OH- = 0.014 mol / 0.150 L = 0.09333 M

Now lets calculate pOH

pOH = -log [ OH-]

pOH = - log [0.09333]

pOH =1.03

using the pOH lets calculate pH

pH + pOH = 14

pH= 14 – pOH

pH = 14 – 1.03

pH = 12.97

Q2) Kc is 1.67*10^20 at 25 C for the formation of iron(III) oxalate complex ion:

Fe^3+ + 3 C2O4^2- ------- > [Fe(C2O4)3]^3-

if .02 M Fe^3+ is initially mixted with 1.00 M oxalate ion, what is the concentration of Fe^3+ ion at equilibrium?

Solution : -

We have the Kc value and the initial concentrations lets make the ICE table

         Fe^3+     +    3 C2O4^2- ------- >    [Fe(C2O4)3]^3-

Lets assume all the Fe^3+ changed to the complex [Fe(C2O4)3]^3-

Therefore assume whatever Fe^3+ remaining in the solution is having concnetration = ‘x’

When 0.02 M Fe^3+ changes to the complex at the same time oxalate ion concentration also decrease by factor 3

That is oxalate ion = 1.0 M –(3 *0.02) = 0.94 M

Now lets use these values in the Kc equation

Kc = [Fe(C2O4)3] / [Fe3+] [C2O4]^3

1.67*10^20 = [0.02] / [x][0.94]^3

Therefore

[x] = [0.02] / 1.67*10^20 * [0.94]^3

X= 1.44*10^-22 M

Therefore equilibrium concentration of the Fe^3+ = 1.44*10^-22 M

Hence option ‘A’ is correct.

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