Respuesta :
Answer: 13.1
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in ml)}}[/tex] .....(1)
Molarity of LiOH solution = 0.350 M
Volume of solution = 50.0 mL
Putting values in equation 1, we get:
[tex]0.350M=\frac{\text{Moles of LiOH}\times 1000}{50.0L}\\\\\text{Moles of LiOH}=\frac{0.350mol/L\times 50.0}{1000}=0.0175mol[/tex]
Moles of [tex]OH^-[/tex] ion = 0.0175 moles
[tex]0.250M=\frac{\text{Moles of}HCLO_4\times 1000}{30.0L}\\\\\text{Moles of} HClO_4=\frac{0.250mol/L\times 25.0}{1000}=0.00625mol[/tex]
Moles of [tex]H^+[/tex] ion = 0.00625 moles
The chemical equation for the reaction of LiOH with [tex]HClO_4[/tex] follows:
[tex]HClO_4+LiOH\rightarrow LiClO_4+H_2O[/tex]
For neutralization:
1 mole of [tex]H^+[/tex] ion will react with 1 mole of [tex]OH^-[/tex] ion
As, 0.00625 moles of [tex]H^+[/tex] ion react with=[tex]\frac{1}{1}\times 0.00625=0.00625[/tex] moles of [tex]OH^-[/tex] ion
Moles of [tex]OH^-[/tex] left = (0.0175-0.00625) = 0.01125 moles
Concentration of [tex]OH^-=\frac{moles}{\text {total volume in L}}=\frac{0.01125}{0.0800L}=0.141M[/tex]
[tex]pOH= -log[OH^-][/tex]
[tex]pOH= -log[0.141]=0.851[/tex]
pH +pOH = 14
pH = 14- pOH = 14 -0.851 = 13.1
Thus pH of the resulting solution is 13.1
The pH of the solution is 13.1.
The equation of the reaction is;
HClO4(aq) + NaOH(aq) -------> NaClO4(aq) + H2O(l)
Number of moles of base= 50/1000 L × 0.350 M = 0.0175 moles
Number of moles of acid= 30/1000 L × 0.250 M = 0.0075 moles
Since the reaction is 1:1, we can see that limiting reactant is the acid HClO4.
Amount of excess base = 0.0175 moles - 0.0075 moles = 0.01 moles
Total volume of solution = 50mL + 30mL = 80mL or 0.08 L
Molarity of excess OH- = 0.01 moles/0.08 L = 0.125 M
Recall that;
pOH = -log[OH-]
pOH = -log[0.125 M]
pOH = 0.9
Again;
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 0.9
pH = 13.1
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