Taking into account the definition of pH and pOH, the mass of LiOH needed to be dissolved in 500.0 mL of water to produce a solution with a pH of 12.40 is 0.299375 grams.
First of all, pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.
The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or H₃O⁺:
pH= - log [H⁺]= - log [H₃O⁺]
Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:
pOH= - log [OH⁻]
The following relationship can be established between pH and pOH:
pOH + pH= 14
In this case, being pH= 12.40, pOH is calculated as:
pOH + 12.40= 14
pOH= 14 - 12.40
pOH= 1.60
Replacing in the definition of pOH the concentration of OH⁻ ions is obtained:
- log [OH⁻]= 1.60
Solving:
[OH⁻]= 10⁻¹ ⁶
[OH⁻]= 0.025 M
Then, the [OH⁻] is 0.025 M.
Strong bases are bases that completely dissociate in water into the cation and OH - (hydroxide ion). LiOH is a strong base, so the OH- concentration is equal to the LiOH concentration:
[OH⁻]= [LiOH]= 0.025 M
Finally, Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume:
[tex]Molarity=\frac{number of moles}{volume}[/tex]
In this case, knowing that molarity is 0.025 M and the volume is 500 mL= 0.5 L, you can replace in the expression for molarity:
[tex]0.025 M=\frac{number of moles}{0.5 L}[/tex]
Solving:
number of moles= 0.025 M× 0.5 L
number of moles= 0.0125 moles
Finally, since the molar mass of LiOH is 23.95 [tex]\frac{g}{mol}[/tex], that is, the amount of mass that the substance contains in one mole, the mass of LiOH that 0.0125 mol contains can be calculated as:
0.0125 mol×23.95 [tex]\frac{g}{mol}[/tex]= 0.299375 g
The mass of LiOH needed to be dissolved in 500.0 mL of water to produce a solution with a pH of 12.40 is 0.299375 grams.
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