Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Q35 E (page 873)

Calculate the concentration of PO43-when Ag3 PO4 starts to precipitate from a solution that is 0.0125M in Ag+.

Short Answer

Expert verified

The value of [PO43-]=9.22×10-16M

Step by step solution

01

Define concentration

The concentration is the abundance of a constituent divided by the total volume of a mixture. Several types of mathematical descriptions can be distinguished: mass concentration, molar concentration, number concentration, and volume concentration.

02

Calculate the concentration

Precipitation of PO43-will begins when the ion product value is higher than solubility product constant value of Ag3 PO4

\[{K_{sp}}=1.8\times{10^{-18}}={\left[{A{g^+}}\right]^3}\left[{PO_4^{3-}}\right]=\left[{PO_4^{3-}}\right]\left({{{0.0125}^3}}\right)\]

\[\left[{PO_4^{3-}}\right]=\frac{{1.8 \times{{10}^{-18}}}}{{{{0.0125}^3}}}=9.22\times{10^{-16}}M\]

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Question: 28. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate \({K_{sp}}\)for each of the slightly soluble solids indicated:

(a) \(AgBr:\left( {A{g^ + }} \right) = 5.7 \times 1{0^{ - 7}}M,\left( {B{r^ - }} \right) = 5.7 \times 1{0^{ - 7}}M\)

(b) \(CaC{O_3}:\left( {C{a^{2 + }}} \right) = 5.3 \times 1{0^{ - 3}}M,\left( {C{O_3}^{2 - }} \right) = 9.0 \times 1{0^{ - 7}}M\)

(c) \(Pb{F_2}:\left( {P{b^{2 + }}} \right) = 2.1 \times 1{0^{ - 3}}M,\left( {{F^ - }} \right) = 4.2 \times 1{0^{ - 3}}M\)

(d) \(A{g_2}Cr{O_4}:\left( {A{g^ + }} \right) = 5.3 \times 1{0^{ - 5}}M,3.2 \times 1{0^{ - 3}}M\)

(e) \(In{F_3}:\left( {I{n^{3 + }}} \right) = 2.3 \times 1{0^{ - 3}}M,\left( {{F^ - }} \right) = 7.0 \times 1{0^{ - 3}}M\)

Question: Using the dissociation constant, \({K_d} = 2.2 \times 1{0^{ - 34}}\), calculate the equilibrium concentrations of\(C{o^{3 + }}\;and\;N{H_3}\)in a\(0.500 - M\;solution of\;Co\left( {N{H_3}} \right)_6^{3 + }\).

Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions.

(a) \(TlCl(s)\) in \(0.025MTlN{O_3}\)

(b) \(Ba{F_2}(\;s)\) in \(0.0313M\;KF\)

(c) \(Mg{C_2}{O_4}\) in \(2.250\;L\)of a solution containing \(8.156\;g\) of \(Mg{\left( {N{O_3}} \right)_2}\)

(d) \(Ca{(OH)_2}(\;s)\) in an unbuffered solution initially with a pH of \(12.700\)

Question: In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. [The reaction of Ag+ with CN– goes to completion, producing the Ag(CN)2 − complex.] Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of Ag(CN)2 −. How many grams of NaCN were in the original sample?

What is the molar solubility of CaF2 in a 0.100 M solution of HF? Kafor HF = 7.2 × 10–4.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free