How many grams of solid potassium hydroxide are needed to exactly neutralize 10.8 mL of a 1.25 M hydrobromic acid solution
KOH + HBr ==> KBr + H2O
mols HBr = M x L = ?
mols KOH = mols HBr (from the coefficients in the equation).
mols KOH = grams/molar mass. You know molar mass and mols, solve for grams.
To determine the number of grams of solid potassium hydroxide (KOH) needed to neutralize the hydrobromic acid (HBr) solution, we need to use the balanced chemical equation between KOH and HBr:
KOH(aq) + HBr(aq) → KBr(aq) + H2O(l)
From the equation, we can see that one mole of KOH reacts with one mole of HBr.
To find the number of moles of HBr, we need to use the given volume and molarity of the hydrobromic acid solution.
Step 1: Calculate the number of moles of HBr.
Given:
Volume of HBr solution = 10.8 mL
Molarity of HBr solution = 1.25 M
Convert the volume of the solution from milliliters (mL) to liters (L):
10.8 mL = 10.8 / 1000 = 0.0108 L
Now, calculate the number of moles of HBr using the formula:
moles = volume (in liters) × molarity
moles of HBr = 0.0108 L × 1.25 mol/L = 0.0135 mol
Step 2: Determine the number of moles of KOH required.
From the balanced chemical equation, we know that one mole of KOH reacts with one mole of HBr. Therefore, the number of moles of KOH required is also 0.0135 mol.
Step 3: Calculate the mass of KOH using its molar mass.
The molar mass of KOH is:
Potassium (K) = 39.1 g/mol
Oxygen (O) = 16.0 g/mol
Hydrogen (H) = 1.0 g/mol
Molar mass of KOH = 39.1 g/mol + 16.0 g/mol + 1.0 g/mol = 56.1 g/mol
Now, calculate the mass of KOH using the formula:
mass = moles × molar mass
mass of KOH = 0.0135 mol × 56.1 g/mol= 0.758 g ≈ 0.76 g
Therefore, approximately 0.76 grams of solid potassium hydroxide (KOH) are needed to exactly neutralize 10.8 mL of a 1.25 M hydrobromic acid solution.