1.000 grams of KHP consumed 43.3 mL of KOH solution. What is the molarity of the KOH solution?
KOH + KHP ==> H2O + K2P.
mols KHP = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols KHP to mols KOH. That's a 1:1 ratio; therefore, ? mols KHP = the same mols KOH.
Then M KOH = mols KOH/L KOH. You know mols and L, solve for M
To find the molarity of the KOH solution, we need to know the number of moles of KHP and the volume of the KOH solution used.
Step 1: Calculate the number of moles of KHP consumed.
The molar mass of KHP (potassium hydrogen phthalate) is 204.22 g/mol.
Given that 1.000 grams of KHP was used, divide this mass by the molar mass to find the number of moles:
1.000 g / 204.22 g/mol = 0.0048926 mol
Step 2: Calculate the molarity of the KOH solution.
The balanced chemical equation between KHP and KOH is:
KHP + KOH → K2HPO4 + H2O
According to the stoichiometry of the equation, the mole ratio between KHP and KOH is 1:1.
Therefore, the number of moles of KOH used is also 0.0048926 mol.
The volume of the KOH solution is given as 43.3 mL, which we will need to convert to liters:
43.3 mL × (1 L / 1000 mL) = 0.0433 L
Now, divide the number of moles of KOH by the volume of the KOH solution in liters to calculate the molarity:
Molarity = Moles / Volume
Molarity = 0.0048926 mol / 0.0433 L ≈ 0.113 M
Thus, the molarity of the KOH solution is approximately 0.113 M.
To find the molarity of the KOH solution, we can use the formula:
Molarity (M) = (moles of solute) / (volume of solution in liters)
First, let's convert the volume of the KOH solution from milliliters (mL) to liters (L):
43.3 mL = 43.3 / 1000 = 0.0433 L
Next, we need to determine the moles of KHP (solute). KHP has a molar mass of 204.23 g/mol:
Number of moles of KHP = mass of KHP / molar mass of KHP
= 1.000 g / 204.23 g/mol
= 0.00489 mol
Since KOH is a 1:1 stoichiometric ratio with KHP, the moles of KOH will be the same. Therefore, the moles of KOH are 0.00489 mol.
Now we can plug the values into the formula to calculate the molarity of the KOH solution:
Molarity = (0.00489 mol) / (0.0433 L)
≈ 0.113 M
Therefore, the molarity of the KOH solution is approximately 0.113 M.