Respuesta :
Answer: The pH of the resulting solution is 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 L)}}[/tex] .....(1)
Molarity of [tex]LiOH[/tex] solution = 0.350 M
Volume of solution = 50.0 mL
Putting values in equation 1, we get:
a) [tex]0.350M=\frac{\text{Moles of} LiOH\times 1000}{50.0ml}\\\\\text{Moles of }LiOH=\frac{0.350mol/L\times 50.0}{1000}=0.0175mol[/tex]
1 mole of [tex]LiOH[/tex] contains = 1 mol of [tex]OH^-[/tex]
Thus [tex]0.0175mol[/tex] of [tex]LiOH[/tex] contain= [tex]\frac{1}{1}\times 0.0175=0.0175[/tex] mol of [tex]OH^-[/tex]
[tex]0.250M=\frac{\text{Moles of} HClO_4\times 1000}{30.0ml}\\\\\text{Moles of }HClO_4=\frac{0.250mol/L\times 30.0}{1000}=0.0075mol[/tex]
1 mole of [tex]HClO_4[/tex] contains = 1 mol of [tex]H^+[/tex]
Thus [tex]0.0075mol[/tex] of [tex]HClO_4[/tex] contain= [tex]\frac{1}{1}\times 0.0075=0.0075[/tex] mol of [tex]H^+[/tex]
[tex]HClO_4+NaOH\rightarrow NaClO_4+H_2O[/tex]
As 1 mole of [tex]H^+[/tex] combines with 1 mole of [tex]OH^-[/tex]
0.0075 moles of [tex]H^+[/tex] combines with = 0.0075 mole of [tex]OH^-[/tex]
Moles of [tex]OH^-[/tex] left = (0.0175-0.0075) = 0.01
Moles of [tex]OH^-[/tex] left = [tex]\frac{moles}{Volume}=\frac{0.01}{80.0}\times 1000=0.125M[/tex]
pOH =[tex]-log[OH^-][/tex]
pOH =[tex]-log[0.125]=0.90[/tex]
[tex]pH+pOH=14[/tex]
[tex]pH=14-0.90=13.1[/tex]
Thus the ph of the resulting solution is 13.1
