Respuesta :
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:
[tex]\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter[/tex]
[tex]\ moles \ in\ F- = 0.100 \ M \times 0.0250 L\\\\[/tex]
[tex]=\ 0.0025 \ moles[/tex]
[tex]\ moles \ in \ HF \ = 0.126M \times 0.0250 L[/tex]
[tex]= 0.00315 \ moles[/tex]
[tex]\ moles \ in \ HCl = 0.0100M \times 0.00500 L[/tex]
[tex]= 0.00005 \ moles[/tex]
[tex]\ Reaction: \\\\F - + H+ \rightarrow HF[/tex]
[tex]\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[/tex]
[tex]\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V[/tex]
[tex]=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M[/tex]
