Question

Which of the following would form a buffer if added to \(250.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{SnF}_{2} ?\) (a) \(0.100 \mathrm{~mol}\) of \(\mathrm{HCl}\) (b) \(0.060 \mathrm{~mol}\) of \(\mathrm{HCl}\) (c) \(0.040 \mathrm{~mol}\) of \(\mathrm{HCl}\) (d) \(0.040 \mathrm{~mol}\) of \(\mathrm{NaOH}\) (e) \(0.040 \mathrm{~mol}\) of \(\mathrm{HF}\)

Short answer

The correct options are: (c) \(0.040 \mathrm{~mol}\) of \(\mathrm{HCl}\) (e) \(0.040 \mathrm{~mol}\) of \(\mathrm{HF}\)

Step-by-step solution

Analyze the reaction between SnF2 and the added substance in each option.

For each of the options, we will analyze if adding the substance forms a conjugate pair with SnF2 (or F-) and if it will maintain the pH of the solution. (a) For 0.100 mol HCl: The reaction will be as follows: SnF2 + 2HCl -> SnCl2 + 2HF Total moles of SnF2: 0.150 M * 0.250 L = 0.0375 mol Total moles of F-: 2 * 0.0375 mol = 0.075 mol As HCl reacts with SnF2 to produce HF, we will not have enough of F- to maintain the pH. Thus, option (a) does not form a buffer. (b) For 0.060 mol HCl: The reaction will be: SnF2 + 2HCl -> SnCl2 + 2HF The moles of SnF2 and F- remain the same as in (a). In this case, we still do not have enough of F- to maintain the pH. Thus, option (b) does not form a buffer. (c) For 0.040 mol HCl: The reaction will be the same as in options (a) and (b). The moles of SnF2 and F- remain the same. We have enough of F- to maintain the pH. Thus, option (c) could form a buffer. (d) For 0.040 mol NaOH: The reaction will be: SnF2 + 2NaOH -> Sn(OH)2 + 2NaF As we can see, NaOH reacts with SnF2 to produce Sn(OH)2 instead of producing a conjugate pair with F-. Thus, option (d) does not form a buffer. (e) For 0.040 mol HF: The reaction will be: SnF2 + HF -> SnF3 + F- Adding HF will result in the formation of the conjugate pair of SnF2 and F-, thus this option could form a buffer.

Conclusion

Based on the analysis of the reactions, options (c) and (e) can form a buffer with 250.0 mL of 0.150 M SnF2. So, the correct options are: (c) \(0.040 \mathrm{~mol}\) of \(\mathrm{HCl}\) (e) \(0.040 \mathrm{~mol}\) of \(\mathrm{HF}\)