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Chapter 4: Q80E (page 224)

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503MAgNO3 requires 20.22 mL of the AgNO3 solution to reach the end point?

AgNO3(aq) + NaCl(aq)鉄禔驳颁濒(s) + NaNO3(aq)

Short Answer

Expert verified

0.3374 M is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503MAgNO3 requires 20.22 mL of the AgNO3 solution to reach the end point.

Step by step solution

01

Given data

This titration is between acid and base. For this formula we use

\({M_a}{V_a} = {M_b}{V_b}\)

Molarity of AgNO3 = 0.2503 M and Volume of Ag NO3 = 20.22 mL = 0.02022 L

Molarity of AgNO3 = M X L = 0.2503 x 0.02022 = 5.061 x 10 -3 mol of Ag NO3

Molarity of NaCl = ? Volume of NaCl = 15 mL = 0.0150 L

02

Determine the concentration

AgNO3(aq) + NaCl(aq)鉄禔驳颁濒(s) + NaNO3(aq)

\(\begin{aligned}{}{M_a}{V_a} = {M_b}{V_b}\\5.061{\rm{ }}x{\rm{ }}{10^{ - 3}}mol{\rm{ }}of{\rm{ }}Ag{\rm{ }}N{O_3}\; \times \frac{{1\,mol\,NaCl}}{{1mol{\rm{ }}of{\rm{ }}Ag{\rm{ }}N{O_3}\;}} = 5.061{\rm{ }}x{\rm{ }}{10^{ - 3}}mol{\rm{ }}\;\end{aligned}\)

Concentration of NaCl

= \(\begin{aligned}{}\frac{{5.061{\rm{ }}x{\rm{ }}{{10}^{ - 3}}mol{\rm{ }}}}{{0.0150L}}\\ = 0.3374\,M\,of\,NaCl\end{aligned}\)

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Most popular questions from this chapter

A sample of solid calcium hydroxide, Ca(OH)2is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with 5.00脳10-2M HCl requires 36.6 mL of the acid to reach the end point.

\({\rm{Ca(OH}}{{\rm{)}}_{2({\rm{aq}})}}{\rm{ + 2HC}}{{\rm{l}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ CaC}}{{\rm{l}}_{2({\rm{aq}})}}{\rm{ + 2}}{{\rm{H}}_2}{{\rm{O}}_{({\rm{l}})}}\)

What is the molarity?

What volume of 0.08892 M HNO3 is required to react completely with 0.2352 g of potassium hydrogen phosphate?

\(2{\rm{HN}}{{\rm{O}}_{3({\rm{aq}})}}{\rm{ + }}{{\rm{K}}_2}{\rm{HP}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ }} \to {\rm{ }}{{\rm{H}}_2}{\rm{P}}{{\rm{O}}_{4({\rm{aq)}}}}{\rm{ + 2KN}}{{\rm{O}}_{3({\rm{aq}})}}\)

Balance the following equations:

\(\begin{array}{l}\;(a)\;PC{l_5}\left( s \right) + {H_2}O\left( l \right) \to POC{l_3}\left( l \right) + HCl\left( {aq} \right)\\\left( b \right)\,Cu\left( s \right) + HN{O_3}\left( {aq} \right) \to Cu{\left( {N{O_3}} \right)_2}\left( {aq} \right) + {H_2}O\left( l \right) + NO\left( g \right)\\\left( c \right){H_2}\left( g \right) + {I_2}\left( g \right) \to HI\left( s \right)\\\left( d \right)\,Fe\left( s \right) + {O_2}\left( g \right) \to F{e_2}{O_3}\left( s \right)\\\left( e \right)\,Na\left( s \right) + {H_2}O\left( l \right) \to NaOH\left( {aq} \right) + {H_2}\left( g \right)\\\left( f \right)\,{\left( {N{H_4}} \right)_2}C{r_2}{O_7}\left( s \right) \to C{r_2}{O_3}\left( s \right) + {N_2}\left( g \right) + {H_2}O\left( g \right)\\\left( g \right){P_4}\left( s \right) + C{l_2}\left( g \right) \to PC{l_3}\left( l \right)\\\left( h \right)PtC{l_4}\left( s \right) \to Pt\left( s \right) + C{l_2}\left( g \right)\end{array}\)

What mass of CO is required to react with 25.13g of Fe2O3 according to the equation

\(F{e_2}{O_3} + CO \to 2Fe + 3C{O_2}\)

How many milliliters of a 0.1500-M solution of KOH will be required to titrate 40.00 mL of a 0.0656-M solution of H3PO4?

\({{\rm{H}}_3}{\rm{P}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + 2KO}}{{\rm{H}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ }}{{\rm{K}}_2}{\rm{HP}}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + 2}}{{\rm{H}}_2}{{\rm{O}}_{({\rm{l}})}}\)
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