Respuesta :
Answer: The concentration of excess [tex][OH^-][/tex] in solution is 0.017 M.
Explanation:
1. [tex]Molarity=\frac{moles}{\text {Volume in L}}[/tex]
moles of [tex]HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles[/tex]
1 mole of [tex]HCl[/tex] give = 1 mole of [tex]H^+[/tex]
Thus 0.019 moles of [tex]HCl[/tex] give = 0.019 mole of [tex]H^+[/tex]
2. moles of [tex]Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles[/tex]
According to stoichiometry:
1 mole of [tex]Ba(OH)_2[/tex] gives = 2 moles of [tex]OH^-[/tex]
Thus 0.012 moles of [tex]Ba(OH)_2[/tex] give = [tex]2 \times 0.012=0.024[/tex] moles of [tex]OH^-[/tex]
[tex]H^++OH^-\rightarrow H_2O[/tex]
As 1 mole of [tex]H^+[/tex] neutralize 1 mole of [tex]OH^-[/tex]
0.019 mole of [tex]H^+[/tex] will neutralize 0.019 mole of [tex]OH^-[/tex]
Thus (0.024-0.019)= 0.005 moles of [tex]OH^-[/tex] will be left.
[tex][OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M[/tex]
Thus molarity of [tex][OH^-][/tex] in solution is 0.017 M.
The concentration of the excess OH ions left in the solution is; M = 0.02 M
We are given;
Molarity of HCl; M = 0.25 M
Volume of HCl; V = 75 mL = 0.075 L
Formula for number of moles is;
n = Molarity × Volume
n_hcl = 0.25 × 0.075
n_hcl = 0.01875 moles
Hydrogen ions in HCl have a valency of +1
Thus; 0.01875 moles of HCL will produce 0.01875 moles of H+
We are given;
Molarity of Ba(OH)2 = 0.055 M
Volume of Ba(OH)2 = 225 mL = 0.225 L
Number of moles of Ba(OH)2 = 0.055 × 0.225 = 0.012375 moles
Ba ions in Ba(OH)2 will have a valency of +2. This means that there will be 2 moles of hydroxyl ions (OH-)
Thus;
0.012375 moles of Ba(OH)2 will produce;
2 × 0.012375 moles of (OH-)
>> 0.02475 moles of (OH-)
Molarity of excess ions left is;
M = (number of moles left)/total volume
Number of moles left = 0.02475 - 0.01875 = 0.006
Total volume = 0.225 + 0.075 = 0.3 L
Thus;
M = 0.06/0.3
M = 0.02 M
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