A 35.0 ml sample of 0.225 m HBr was titrated with 42.3 ml of KOH. What is the concentration of the KOH
For simple 1:1 equations, one may use
mL acid x M acid = mL base x M base.
To determine the concentration of the KOH, we can use the balanced chemical equation for the reaction between HBr and KOH:
HBr + KOH -> KBr + H2O
From the balanced equation, we can see that the stoichiometric ratio between HBr and KOH is 1:1. This means that for every mole of HBr reacted, we will have 1 mole of KOH reacted.
Let's calculate the number of moles of HBr in the solution:
Volume of HBr solution = 35.0 ml
Concentration of HBr = 0.225 mol/L
Number of moles of HBr = Volume x Concentration = 35.0 ml x 0.225 mol/L = 7.875 mmol (millimoles)
Since the stoichiometric ratio between HBr and KOH is 1:1, this means that 7.875 mmol of KOH also reacted with HBr.
Now, let's calculate the concentration of KOH:
Volume of KOH solution = 42.3 ml
Number of moles of KOH = 7.875 mmol (since the stoichiometric ratio is 1:1)
Concentration of KOH = Number of moles / Volume = 7.875 mmol / 42.3 ml = 0.186 M
The concentration of KOH is approximately 0.186 M.
To determine the concentration of KOH, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between HBr and KOH:
HBr + KOH → KBr + H2O
From the balanced equation, we can see that one mole of HBr reacts with one mole of KOH to produce one mole of KBr and one mole of water.
First, let's calculate the number of moles of HBr:
Moles of HBr = (volume of HBr solution in liters) x (concentration of HBr in moles per liter)
= 0.0350 L x 0.225 mol/L
= 0.007875 mol
Since the reaction is 1:1 between HBr and KOH, the number of moles of KOH used is equal to the number of moles of HBr. Therefore, the number of moles of KOH is also 0.007875 mol.
Next, we can calculate the concentration of KOH:
Concentration of KOH = (moles of KOH) / (volume of KOH solution in liters)
= 0.007875 mol / 0.0423 L
≈ 0.186 mol/L
Therefore, the concentration of the KOH solution is approximately 0.186 mol/L.