Question

What is the PH of a solution made by mixing 30.00 mL of 0.10 M HCI with 40.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. A) 12.15 B) 13.15 C) 1.85 D) 0.85 28) Which of the following titration result in a basic soltuion at the equilvalence point? 29) The balanced equation for the solubility equilibrium of Fe (OH)_2 is shown below. What is the equilibrium constant expression for the K_sp of (OH)_2?

Answer 1

27.

First we need to obtain the moles for each compound

$30 mL\frac{0.10 mol HCl}{1000 mL}=3x10^{-3}moles HCl$

$40 mL\frac{0.10 mol KOH}{1000 mL}=4x10^{-3}moles KOH$

Then we write the process

	HCl	KOH
Initial	0	4x10-3 mol
Change	3x10-3 mol	---
Eq	0	1x10-3 mol

Then dividing the moles of KOH between the total volume:

$\frac{1x10^{-3}}{70 mL}\frac{1000 mL}{L}=0.01429 M KOH$

Using the following equation we get:

$pOH=-log[OH]$

$pOH=-log[0.01429 M]$

$pOH=1.845$

Using the relation between pOH and pH

$pH+pOH=14$

$pH=14-1.845$

$pH=12.15$

The answer is A) 12.15

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