Chapter 23: Q14P (page 629)

23-14 For the extraction of ${Cu}^{2 +}$ by dithizone ${CCI}_{4}, K_{L} = 1.1 \times 10^{4}, K_{M} = 7 \times 10^{4,} K_{a} = 3 \times 10^{- 5}, b = 5 \times 10^{22}, n = 2$
(a) Calculate the distribution coefficient for extraction of 0.1mM ${Cu}^{2 +}$
into ${CCI}_{4}$ by 0.1mM dithizone at and at pH = 1 and at pH = 4 .
(b) If 100mL of 0.1mM aqueous ${Cu}^{2 +}$ are extracted once with 10mL
of 0.1mM dithizone atpH = 1 , what fraction of ${Cu}^{2 +}$ remains in the aqueous phase?

Short Answer

Expert verified

$a) At pH = 4 D = 2.6 \times 10^{10}, D = 2.6 \times 10^{4} b) At pH = 1 q = 3.8 \times 10^{- 4}$

Step by step solution

Given Information:

$K_{L} = 1.1 \times 10^{4}, K_{M} = 7 \times 10^{4,} Ka = 3 \times 10^{- 5}, b = 5 \times 10^{22}, n = 2$

To Find the Distribution coefficients:

For the distribution coefficients of $0.1 μM {Cu}^{2 +}$ into ${CCI}_{4}$ by 0.1mM dithizone at pH = 1 and at pH = 4

localid="1654779544284" $D = \frac{K_{M} {βK}_{a}^{n}}{K_{L}^{n}} \frac{{[HL]}_{org}^{n}}{{[H^{+}]}_{aq}^{n}} [HL] = 0.1 mM = 1 \times 10^{- 4} M D = \frac{(7 \times 10^{4}) (5 \times 10^{22}) {(3 \times 10^{- 5})}^{2}}{{(1.1 \times 10^{4})}^{2}} \frac{{(1 \times 10^{- 4})}^{2}}{{[H^{+}]}^{2}}$

To calculate $[H^{+}] = 10^{- pH} = 10^{- 4}$

$D = \frac{(7 \times 10^{4}) (5 \times 10^{22}) {(3 \times 10^{- 5})}^{2}}{{(1.1 \times 10^{4})}^{2}} \frac{{(1 \times 10^{- 4})}^{2}}{{[10^{- 4}]}^{2}} D = 2.6 \times 10^{10} For pH = 1 D = \frac{K_{M} {βK}_{a}^{n}}{K_{L}^{n}} \frac{{[HL]}_{org}^{n}}{{[H^{+}]}_{aq}^{n}} [HL] = 0.1 mM = 1 \times 10^{- 1} M D = \frac{(7 \times 10^{4}) (5 \times 10^{22}) {(3 \times 10^{- 5})}^{2}}{{(1.1 \times 10^{4})}^{2}} \frac{{(1 \times 10^{- 4})}^{2}}{{[H^{+}]}^{2}} [H^{+}] = 10^{- pH} = 10^{- 1} D = \frac{(7 \times 10^{4}) (5 \times 10^{22}) {(3 \times 10^{- 5})}^{2}}{{(1.1 \times 10^{4})}^{2}} \frac{{(1 \times 10^{- 1})}^{2}}{{[10^{- 1}]}^{2}} D = 2.6 \times 10^{4}$

To find the aqueous phase :

To calculate the fraction of ${Cu}^{2 +}$ in aqueous phase if 100mL of $0.1 μM$ aqueous phase ${Cu}^{2 +}$ 10mM dithizone at pH = 1.0

q = \frac{V_{aqueous {Cu}^{2 +}}}{{V_{aqueous {Cu}^{2 +}}}^{+} (D \times V_{dithizione})} q = \frac{100 mL}{100 mL + ((2.6 \times 10^{4}) \times 10 mL)} q = 3.8 \times 10^{- 4}

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Short Answer

Step by step solution

Given Information:

To Find the Distribution coefficients:

To find the aqueous phase :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Chemical Reactions

Kinetics

Ionic and Molecular Compounds

Chemistry Branches

Physical Chemistry

Study anywhere. Anytime. Across all devices.