/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q79E Question: In a titration of cyan... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Q79E (page 877)

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鈥� goes to completion, producing the Ag(CN)2 鈭� 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 鈭�. How many grams of NaCN were in the original sample?

Short Answer

Expert verified

\(2.815 \cdot {10^{ - 2}}{\rm{g}}\)of NaCN were in the original sample.

Step by step solution

01

Calculate the number of moles of Ag+-:

The reaction

\({\rm{A}}{{\rm{g}}^ + } + 2{\rm{C}}{{\rm{N}}^ - } \to {\rm{Ag}}({\rm{CN}})_2^ - \)

  • 28.72 mL of 0.0100 M AgNO3

Let us calculate the mass of NaCN that was in the original sample.

Calculate the number of moles of AgNO3 is

\(\begin{array}{*{20}{c}}{{n_{{\rm{AgN}}{{\rm{O}}_3}}} = 0.0100{\rm{M}} \cdot 28.72{\rm{mL}} \cdot \frac{{1{\rm{L}}}}{{1000{\rm{mL}}}}}\\{ = 2.872 \cdot {{10}^{ - 4}}{\rm{mol}}}\end{array}\)

Since 1 mole of AgNO3 dissociates completely into 1 mole of Ag+ and 1 mole of NO3- , the number of moles of Ag+is\(2.872 \cdot {10^{ - 4}}{\rm{mol}}\).

02

Calculate the mass of NaCN:

  • Now let us find the number of moles of NaCN that was in the original sample.

Since 1 mole of Ag+reacts with 2 moles of CN-, the number of moles of CN- is

\(\begin{array}{*{20}{c}}{{n_{C{N^ - }}} = 2 \cdot {n_{{\rm{A}}{{\rm{g}}^ + }}}}\\{ = 2 \cdot 2.872 \cdot {{10}^{ - 4}}{\rm{mol}}}\\{ = 5.744 \cdot {{10}^{ - 4}}{\rm{mol}}}\end{array}\)

Therefore, the number of moles of NaCN is\(5.744 \cdot {10^{ - 4}}{\rm{mol}}\).

  • Finally, let us calculate the mass of NaCN

\(\begin{array}{*{20}{c}}{{m_{NaCN}} = {n_{NaCN}} \cdot {M_{NaCN}}}\\{ = 5.744 \cdot {{10}^{ - 4}}{\rm{mol}} \cdot 49.0072{\rm{g}}/{\rm{mol}}}\\{ = 2.815 \cdot {{10}^{ - 2}}{\rm{g}}}\end{array}\)

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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: Using the dissociation constant, \({K_d} = 1 \times 1{0^{ - 44}}\), calculate the equilibrium concentrations of \(F{e^{3 + }}\;and\;C{N^ - }\) in a \(0.333M\) solution of \(Fe(CN)_6^{3 - }\).

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

Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products):

(a) \(AgI\)

(b) \(A{g_2}S{O_4}\)

(c) \(Mn{(\;OH\;)_2}\)

(d) \(Sr{(\;OH\;)_2} \times 8{H_2}O\)

(e) The mineral brucite, \(Mg{(\;OH\;)_2}\)

Question: Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from seawater by the following sequence of reactions:

\(\begin{array}{*{20}{c}}{M{g^{2 + }}(aq) + Ca{{(OH)}_2}(aq) \to Mg{{(OH)}_2}(s) + C{a^{2 + }}(aq)}\\{Mg{{(OH)}_2}(s) + 2HCl(aq) \to MgC{l_2}(s) + 2{H_2}O(l)}\end{array}\)

\(MgC{l_2}(l)\mathop \to \limits^{\;electrolysis\;} Mg(s) + C{l_2}(g)\)

Sea water has a density of 1.026 g/cm3 and contains 1272 parts per million of magnesium \(M{g^{2 + }}(aq)\)by mass. What mass, in kilograms, \(Ca{(OH)_2}\)is required to precipitate 99.9% of the magnesium in 1.00 脳 103 L of seawater?

Calculate the molar solubility of \({\bf{CdC}}{{\bf{O}}_{\bf{3}}}\) in a buffer solution containing \({\bf{0}}.{\bf{115}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{N}}{{\bf{a}}_{\bf{2}}}{\bf{C}}{{\bf{O}}_{\bf{3}}}{\rm{ }}{\bf{and}}{\rm{ }}{\bf{0}}.{\bf{120}}{\rm{ }}{\bf{M}}{\rm{ }}{\bf{NaHC}}{{\bf{O}}_{\bf{3}}}\) .
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

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.