Question

What is the pH of a solution prepared by mixing 100.00 mL of 0.020 M Ca(OH)2 with 50.00 mL of 0.100 M NaOH? Assume that the volumes are additive?a. 12.78.b. 13.25.c. 12.67.d. 12.95.

Answer 1

Answer:

The pH of the solution is 12.78.

Explanation:

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution. The pOH is defined as the negative logarithm of the activity of the hydroxide ions. That is, the concentration of OH- ions:

pOH= - log [OH-]

On the other side, Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume. Molarity is calculated as:

$Molarity= \frac{number of moles}{volume}$

Molarity is expressed in units: $\frac{moles}{liter}$

In this case, the solution is prepared by mixing 100 ml (equal to 0.1 L, where 1000 mL = 1 L) of Ca(OH)₂ 0.020 M with 50 ml (equal to 0.05 L) of 0.100 M NaOH. Then, Ca(OH)₂ and NaOH are strong bases, so they dissociate completely. In the case of the first hydroxide, for each mole of Ca(OH)₂,

form two moles of OH-. In the case of sodium hydroxide, for each mole of hydroxide, one mole of OH- is formed. So, taking into account the definition of molarity, the number of moles of OH- that each hydroxide contributes to the solution is calculated as:

From Ca(OH)₂: 0.1 L* 0.02 M*2 = 0.004 moles

From NaOH: 0.05 L* 0.1 M= 0.005 moles

So, the amount of total moles of OH- is the sum that each hydroxide contributes to the solution: 0.004 moles + 0.005 moles= 0.009 moles

On the other hand, volumes are additive. Then: 0.1 L +  0.05 L= 0.15 L

Replacing in the definition of molarity the number of moles and the volume:

$[OH-]=\frac{0.009 moles}{0.15 L}$

Solving:

[OH-]= 0.06 $\frac{moles}{liter}$

Replacing in the definition of pOH:

pOH= - log 0.06

pOH= 1.22

The following relationship can be established between pH and pOH:

pH + pOH= 14

Being pOH= 1.22 and replacing:

pH + 1.22= 14

pH= 14 - 1.22

pH= 12.78

The pH of the solution is 12.78.