Question

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 5.00 mL of 0.0100 M HCl to 25.0 mL of this solution is __________ M.
a. 0.0980
b. 0.0817
c. 0.0167
d. 0.0850
e. 0.00253

Answer 1

Answer:

The answer is "Option b"

Explanation:

In this question first we calculates the moles in F-, HF, and in HCL, which can be defined as follows:

Formula:

$\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter$

$\ moles \ in\ F- = 0.100 \ M \times 0.0250 L$

                     $=\ 0.0025 \ moles$

$\ moles \ in \ HF \ = 0.126M \times 0.0250 L$

                       $= 0.00315 \ moles$

$\ moles \ in \ HCl = 0.0100M \times 0.00500 L$

                       $= 0.00005 \ moles$

$\ Reaction: \\\\F - + H+ \rightarrow HF$

$\Rightarrow \ moles \ in \ F- = 0.0025 \\\\\Rightarrow \ moles \ in \ H+ = 0.00005 \\\\ \Rightarrow \ moles \ in \ HF = 0.00315\\\\ \ total \ moles = 0.00250 -0.0000500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00315 + 0.00005\\\\\ total \ moles =0.00245 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00245$

$\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V$

                                                                                $=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M$