Question #c9dab

So, you know that you've got a solution that contains fluoride ions, #F^(-)#, and hydrogen fluoride, #HF#.

The key to solving this problem is the reaction that takes place when hydrogen chloride is added to the solution. The fluoride ions will react with the hydrogen chloride to produce hydrogen fluoride

#H_((aq))^(+) + F_((aq))^(-) -> HF_((aq))#

To see what goes on when this reaction takes place you have to work with moles. Use the molarities of the solutions you're adding together to determine how many moles of each species you have

#C = n/V => n = C/V#

#n_(F^(-)) = "0.100 M" * 25 * 10^(-3)"L" = "0.0025 moles F"^(-)#

#n_(HF) = "0.126 M" * 25 * 10^(-3)"L" = "0.00315 moles HF"#

#n_(HCl) = "0.0100 M" * 5 * 10^(-3)"L" = "0.00005 moles HCl"#

So, when you add the two solutions together, the hydrochloric acid will be consumed. At the same time, the number of moles of fluoride ions will decrease by the number of moles of #HCl# that reacted.

Moreover, the number of moles of #HF# will increase by the number of moles of #HCl# that treacted.. This means that you'll get

#n_(HCl) = 0#

#n_(F^(-)) = 0.0025 - 0.00005 = "0.002495 moles F"^(-)#

#n_(HF) = 0.00315 + 0.00005 = "0.0032 moles HF"#

The total volume of the solution will be

#V_"sol" = V_1 + V_(HCl) = 25.0 + 5.00 = "30.0 mL"#

As a result, the concentration of the hydrogen fluoride will be

#[HF] = "0.0032 moles"/(30.0 * 10^(-3)"L") = "0.10667 M"#

Rounded to three sig figs, the answer will indeed be

#[HF] = color(green)("0.107 M")#