6. (2 points) 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, assume the reaction goes to completion.

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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 13.1

Explanation:

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

    .....(1)

Molarity of LiOH solution = 0.350 M

Volume of solution = 50.0 mL

Putting values in equation 1, we get:

Moles of ion = 0.0175 moles

Moles of ion = 0.00625 moles

The chemical equation for the reaction of LiOH with follows:

For neutralization:

1 mole of ion will react with 1 mole of ion

As, 0.00625 moles of ion react with= moles of ion

Moles of left = (0.0175-0.00625) = 0.01125 moles

Concentration of

pH +pOH = 14

pH = 14- pOH = 14 -0.851 = 13.1

Thus pH of the resulting solution is 13.1