If 45.1 mL of 0.112 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?
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We can use the balanced equation between HCl and KOH to solve this problem. The balanced equation is:
HCl + KOH → KCl + H2O
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of KOH. Therefore, the number of moles of HCl can be determined using the formula:
moles = concentration × volume
Given: concentration of HCl = 0.112 M, volume of HCl = 45.1 mL
Converting the volume to liters:
volume = 45.1 mL × (1 L / 1000 mL) = 0.0451 L
Using the formula:
moles of HCl = 0.112 M × 0.0451 L = 0.0050512 moles
Since the balanced equation shows a 1:1 mole ratio between HCl and KOH, the number of moles of KOH must also be 0.0050512 moles.
The molar mass of KOH is K (39.1 g/mol) + O (16.0 g/mol) + H (1.0 g/mol) = 56.1 g/mol.
Using the formula:
mass = moles × molar mass
mass of KOH = 0.0050512 moles × 56.1 g/mol = 0.283 g
Therefore, there must be 0.283 grams of KOH present in the solution.
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