Answer to Question #117376 in General Chemistry for Jihye Yu

Question

(all work above must be complete before attempting the bonus) You are stuck with a problem. You need to measure pH of a solution known to be made from a metal hydroxide, but you don’t have a meter or any indicators. You do happen to have some lead (II) nitrate that is soluble, and you remember that lead (II) hydroxide is insoluble. You add some to 1 liter of your unknown solution and a precipitate forms. You add more until the precipitate stops forming and then a bit more just in case. After you filter and dry the precipitate, you have 3.81 grams of it. What was the approximate pH of the original solution? SHOW YOUR STEPS AND YOUR WORK. EXPLAIN EACH STEP SO I KNOW WHAT YOU DID

Answer 1

You have to assume your unknown hydroxide is completely dissociate in aqueous solution. If so, then you can calculate pOH

pOH of the solution as pOH=−log[OH

−

]

pOH=−log⁡[OHX−]. Then, pH=pK

w

−pOH=pK

w

+log

[OH

−

]

pH=pKw−pOH=pKw+log⁡[OHX−]. Now, you see, since pK

w

=14

pKw=14, only unknown is [OH

−

]

[OHX−].

Now look at the reaction of your unknown hydroxide with lead (II) nitrate:

2OH

−

(aq)+Pb(NO

3

)

2

(aq)

⟶Pb(OH)

2

(s)+2NO

3

(aq)

2OHX−(aq)+Pb(NOX3)X2(aq)⟶Pb(OH)X2(s)+2NOX3(aq)

Since you know the precipitated amount of Pb(OH)

2

Pb(OH)X2 isolated from 1.0 L

1.0 L of unknown hydroxide solution, you can calculate original [OH

−

]

[OHX−]:

Amount of OH

−

 in original solution

=3.81 g of Pb(OH)

2

×1 mol of Pb(OH)

2

241.21 g of Pb(OH)

2

×2 mol of OH

−

1 mol of Pb(OH)

2

=0.0316 mol of OH

−

Amount of OHX− in original solution=3.81 g of Pb(OH)X2×1 mol of Pb(OH)X2241.21 g of Pb(OH)X2×2 mol of OHX−1 mol of Pb(OH)X2=0.0316 mol of OHX−

Thus,[OH

−

]=0.0316 M

[OHX−]=0.0316 M, and hence,pH=pK

w

+log[OH

−

]=14+log

(0.0316)=12.5

pH=pKw+log⁡[OHX−]=14+log⁡(0.0316)=12.5

Asking is approximate pH

pH of orignal solution. Thus, this answer is acceptable. However, keep in mind that trace of OH

−

OHX− will remain in solution due to the solubility of Pb(OH)

2

Pb(OH)X2 in water (K

sp

=1.42×10

−20

)

(Ksp=1.42×10−20).