Question

Please show all the work that you did to come to your conclusion. Thank you

Question 1 This is a multiple-choice question, but you need to show your work to support your choice. Which of the following would form a buffer if added to 250.0 mL of 0.150 M SnF2? (a) 0.100 mol of HCI (b) 0.060 mol of HCI (c) 0.040 mol of NaOH

Answer 1

Given, Volume of SnF₂ = 250.0 mL

Molarity of SnF₂ = 0.150 M

No. of moles of SnF₂ = Molarity x Volume (L) = Molarity x Volume (mL)/1000

= 0.150x250/1000 = 0.0375 mol

We know that SnF₂ is a salt to give two F- in the water solution and reacts with the strong acid to produce the weak acid HF. The equation is given below:

2HCl + SnF₂ ----> 2HF + SnCl₂

From the above equation, we can say that 2 mole of HCl reacts with one mole of SnF2 to form 2 mol of HF. We know for making a buffer solution, we need some amount of weak acid (HF) and some amount of their conjugate base (F^-) in the solution. So, we have to select the answer where some amount of SnF₂ left unreacted i.e. some F- is left in the reaction to make the buffer.

Here,

Buffer: A weak acid, HF + its conjugate base, F^-

option (a)

Moles of SnF₂ = 0.0375 (calculated)

Given,

moles HCl = 0.100

From the above balance equation, 0.100 moles of HCl react with 0.050 moles of SnF₂ to give 0.050 moles of HF. So, in this case, 0.0375 moles SnF₂ is a limiting reagent and completely react with HCl and form HF. Hence there is no SnF2 or F^- left in the reaction.

That's why this is not a buffer solution.

option (b)

Moles of HCl = 0.060

Moles of SnF₂ = 0.0375

Here, 0.060 moles of HCl react with 0.030 moles of SnF₂ to form 0.030 moles of HF. So, limiting reagent, in this case, is HCl. Hence, 0.0075 moles of SnF₂ left i.e. (2x0.0075) 0.0150 mole of F^- left.

This solution is called as a buffer solution.

Option (b) is a correct answer.