How many milliliters of a 0.20 M KOH are needed to completely neutralize 90.0 milliliters of 0.10 M HCl?
mL x M = mL x M
i’m thinking possibly 90, but it could be wrong
To find the number of milliliters of KOH needed to neutralize a given volume of HCl, we can use the concept of molar stoichiometry and the balanced chemical equation for the reaction between KOH and HCl.
The balanced chemical equation for the reaction between KOH and HCl is:
KOH + HCl → KCl + H₂O
From the balanced equation, we can see that one mole of KOH reacts with one mole of HCl to produce one mole of water (H₂O). This means that the number of moles of KOH required to neutralize the given volume of HCl is equal to the number of moles of HCl.
First, we need to calculate the number of moles of HCl using the formula:
moles = concentration (M) × volume (L)
Given that the volume of HCl is 90.0 milliliters and the concentration is 0.10 M, we can convert the volume to liters:
volume (L) = volume (mL) ÷ 1000
volume (L) = 90.0 mL ÷ 1000
volume (L) = 0.090 L
Now we can calculate the number of moles of HCl:
moles HCl = concentration HCl × volume HCl
moles HCl = 0.10 M × 0.090 L
moles HCl = 0.009 moles
Since the stoichiometry of KOH to HCl is 1:1, we know that 0.009 moles of KOH will be needed.
Finally, we can calculate the volume of 0.20 M KOH required using the formula:
volume (L) = moles ÷ concentration (M)
volume KOH = 0.009 moles ÷ 0.20 M
volume KOH = 0.045 L
To convert the volume to milliliters, we can multiply by 1000:
volume KOH = 0.045 L × 1000
volume KOH = 45.0 mL
Therefore, 45.0 milliliters of 0.20 M KOH are needed to completely neutralize 90.0 milliliters of 0.10 M HCl.