Suppose 50.0 ml of 0.350 m lithium hydroxide is mixed with 30.0 ml of 0.250 m perchloric acid. What is the ph of the resulting solution

Question

asked May 23, 2021 89.5k views

1 Answer

Amit Patil · Answer 1 · 2021-05-27T20:57:10+0000

Answer: The pH of the resulting solution is 13.1

Step-by-step explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}* 1000}{\text{Volume of solution (in L)}}$ .....(1)

Molarity of
LiOH solution = 0.350 M

Volume of solution = 50.0 mL

Putting values in equation 1, we get:

a)
$0.350M=\frac{\text{Moles of} LiOH* 1000}{50.0ml}\\\\\text{Moles of }LiOH=(0.350mol/L* 50.0)/(1000)=0.0175mol$

1 mole of
LiOH contains = 1 mol of
OH^-

Thus
0.0175mol of
LiOH contain=
(1)/(1)* 0.0175=0.0175 mol of
OH^-

$0.250M=\frac{\text{Moles of} HClO_4* 1000}{30.0ml}\\\\\text{Moles of }HClO_4=(0.250mol/L* 30.0)/(1000)=0.0075mol$

1 mole of
HClO_4 contains = 1 mol of
H^+

Thus
0.0075mol of
HClO_4 contain=
(1)/(1)* 0.0075=0.0075 mol of
H^+

$HClO_4+NaOH\rightarrow NaClO_4+H_2O$

As 1 mole of
H^+ combines with 1 mole of
OH^-

0.0075 moles of
H^+ combines with = 0.0075 mole of
OH^-

Moles of
OH^- left = (0.0175-0.0075) = 0.01

Moles of
OH^- left =
(moles)/(Volume)=(0.01)/(80.0)* 1000=0.125M

pOH =
-log[OH^-]

pOH =
-log[0.125]=0.90

pH+pOH=14

pH=14-0.90=13.1

Thus the ph of the resulting solution is 13.1