A beaker contains 100 mL of water. We add 112 mg of KOH(s). What is the concentration of H+ions?
mols KOH = grams/molar mass
M KOH = mols/L soln.
M OH^- = M KOH
pOH = -log(OH^-) and go from there.
To determine the concentration of H+ ions, we need to calculate the moles of KOH(s) added and then use the stoichiometry of the reaction to find the moles of H+ ions.
First, let's convert the mass of KOH(s) to moles. The molar mass of KOH is 56.1 g/mol.
Given:
Mass of KOH(s) = 112 mg = 112 × 10^(-3) g
Molar mass of KOH = 56.1 g/mol
Now, we can calculate the moles of KOH:
Moles of KOH = Mass of KOH / Molar mass of KOH
= 112 × 10^(-3) g / 56.1 g/mol
≈ 0.002 mol
The balanced chemical equation for the reaction between KOH and water is:
KOH + H2O → K+ + OH- + H2O
From the equation, we can see that for every 1 mole of KOH, we will have 1 mole of H+ ions.
Thus, the moles of H+ ions present in the solution is also 0.002 mol.
To determine the concentration of H+ ions, we need to know the volume of the solution. You mentioned that the beaker contains 100 mL of water. Assuming that the volume remains 100 mL after adding KOH, we can now calculate the concentration.
Concentration (C) = Moles / Volume
= 0.002 mol / 100 mL
= 0.002 mol / 0.1 L
= 0.02 mol/L
So, the concentration of H+ ions in the solution is 0.02 mol/L.