Question

(e) Consider this example problem: If 100 mL of 0.100 M HCl solution is mixed with 100. mL of 0.100 M NaOH, what is the molarity of the resulting salt solution? (assuming the volumes are additive and ignore the change in H2O, which is negligible). HCIA NaOHa NaCl + H2O 1 mol 1 mol 1 mol Mols @Start: 100 mL (100 M ME!) 100 ml (0.1.000 ) O mol = 10 mmol HBr = 10 mmol NaOH Change - 10 mmol - 10 mmol NaOH + 10 mmol O mmol 0 mmol 10 mmol Rxn ratio: 1 mol After rxn The resulting solution contains 10 mmol NaCl in a total volume of 200 ml. The molarity of NaCl solution is 10 mmol - = 0.05 M NaCl 200 ml Fill in a table like this for the calculations in problem (1) (a)-(d). (2) Now, show the calculations for the same reaction (HBr + NaOH) if 100 ml of 0.250 M HBr and 200.0 mL of 0.250 M NaOH are mixed. Determine the molarities of the solutes in the resulting solution.

Answer 1

$HBr+NaOH \to NaBr+H_2O$

	HBr	NaOH	NaBr
Reaction ratio	1 mol	1 mol	1 mol
Moles at start	0.250 тmd 100. тLx 1 mL — 25 тmol	$200. \ mL \times \frac{0.250 \ mmol}{1 \ mL} \\ \ \\ \ =50 \ mmol$
Change
After reaction

Total volume

The molarity of NaBr solution is $\frac{25 \ mmol}{ 300. \ mL}=0.0833 \ M \ NaBr$

Molarity of NaOH solution is $\frac{25 \ mmol}{ 300. \ mL}=0.0833 \ M \ NaOH$