Respuesta :
Explanation:
[tex]HBr(aq) + LiOH(aq) \rightarrow LiBr(aq)+H_2O(l)[/tex]
a)Molarity of LiOH,[tex]M_1[/tex] = 0.0024 M
Volume of LiOH = [tex]V_1[/tex]
Molarity of HBr,[tex]M_2[/tex] = 0.0026 M
Volume of HBr = [tex]V_2=22mL=[/tex]
According to reaction , 1 mol of LiOH neutralizes 1 mol of HBr.
[tex]M_1V_1=M_2V_2[/tex]
[tex]V_1=\frac{0.0026 M\times 22mL}{0.0024 M}=23.8333 mL[/tex]
[tex]M_1=\frac{\text{Molesof LiOH}}{V_1}[/tex]
Moles of LiOH = [tex]0.0024 mol/L\times 23.8333 ml=0.0571 mol[/tex]
Mass of 0.0571 mol of LiOH:
[tex]0.0571 mol\times 2 g/mol=1.3704 g[/tex]
1.3704 g of 0.0024 M LiOH that would neutralize 22 mL of 0.0026 M HBr.
b) According to reaction , 1 mol LiOH gives 1 mol of LiBr.
Then ,0.0571 mol of LiOH will give:
[tex]\frac{1}{1}\times 0.0571 mol=0.0571 mol[/tex] of LiBr
0.0571 moles of salt are produced in the reaction
c) Moles of salt = 0.0571 mol
Volume of the solution = 22 ml+ 23.8333 mL= 45.8333 mL = 0.0458333 L
Molar concentration of the salt:
[tex]{LiBr}=\frac{0.0571 mol}{0.0458333 L}=1.2458 mol/L[/tex]
1.2458 mol/L is the molar concentration of the salt after the reaction is complete.
