A 300 mL solution of 0.0500 M HCl is titrated with 1.00 M KOH. How many mL of the KOH solution will be needed to reach the equivalence point of this titration?
A.5 mL
B.10 mL
C.50 mL
D20 mL
E15 mL
The answer i got was E. 15ml
Can you explain to me why?
HCl + KOH ==> KCl + H2O
mols HCl = M x L = 0.015
mols KOH = the same since 1 mol HCl reacts with 1 mol KOH = 0.015
Then M KOH = mols KOH/ L KOH. You know M and mols, solve for L.
1.00 = 0.015/L and L = 0.015/1.00 = 0.015L. Convert that to mL for 15 mL.
To find the number of mL of the KOH solution needed to reach the equivalence point of this titration, we need to use the balanced chemical equation for the reaction between HCl and KOH:
HCl + KOH -> H2O + KCl
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of KOH. This means that the number of moles of KOH required to neutralize the HCl is the same as the number of moles of HCl.
First, let's calculate the number of moles of HCl in the solution given the volume and concentration:
Moles of HCl = volume (in L) x concentration (in mol/L)
Moles of HCl = 300 mL x (1 L/1000 mL) x 0.0500 mol/L
Moles of HCl = 0.015 mol
Since the number of moles of KOH required is the same as the number of moles of HCl, we can calculate the volume of the KOH solution needed:
Volume of KOH solution (in L) = moles of KOH / concentration of KOH (in mol/L)
Volume of KOH solution = 0.015 mol / 1.00 mol/L
Volume of KOH solution = 0.015 L = 15 mL
Therefore, the correct answer is E. 15 mL.