Q81E - Chemistry (OpenStax)

Question

In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO3)2 solution.

2Cl−(aq) + Hg(NO₃)₂(aq) ⟶ 2NO₃⁻(aq) + HgCl₂(s)

 

What is the Cl⁻ concentration in a 0.25-mL sample of normal serum that requires 1.46 mL of 8.25 × 10⁻⁴ M Hg(NO₃)₂(aq) to reach the end point?

Step-by-Step Solution

Answer

The required Cl⁻ concentration is 9.6 x 10^-3 M 

Given Data

2Cl⁻(aq) + Hg(NO₃)₂(aq) ⟶ 2NO₃⁻(aq) + HgCl₂(s)

 

Hg(NO₃)₂(aq)   =   8.25 × 10⁻⁴ M

 

Volume of Hg(NO₃)₂  =  1.46 mL   =  0.01146 L

Sample volume = 0.25 mL = 2.5 x 10 ^-4 L

 

Molarity(M) = mol/ L  

 

So, no of mol of Hg(NO₃)₂ =   M  x L  

                                           =   8.25 × 10⁻⁴ M  x   0.00146 L

   

                                            =1.2045 x 10^-6 Hg(NO₃)₂

 

Determine concentration of Cl −

\(\begin{aligned}{\underline{\phantom{xx}}}\frac{{1.2045{\rm{ }}x{\rm{ }}{{10}^{ - 6}}Hg{{\left( {N{O_3}} \right)}_2} \times 2\,mol\,of\,C{l^ - }}}{{1\,mol\,Hg{{\left( {N{O_3}} \right)}_2}}}\\ = 2.409 \times {10^{ - 6}}\,mol\,of\,C{l^ - }\end{aligned}\)

Required concentration of Cl         

\(\begin{aligned}{\underline{\phantom{xx}}} = \frac{{2.409 \times {{10}^{ - 6}}\,mol\,of\,C{l^ - }}}{{2.5 \times {{10}^{ - 4}}L}}\\ = 9.6 \times {10^{ - 3}}M\end{aligned}\)