The complete balanced chemical equation for this is:
<span>3KOH + H3PO4
--> K3PO4 + 3H2O</span>
First we calculate the number of moles of H3PO4:
moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol
From stoichiometry, 3 moles of KOH is required for every
mole of H3PO4, therefore:
moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole
H3PO4) = 0.0468 mol
Calculating for volume given molarity of 0.350 M KOH:
Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7
mL
Answer:
<span>133.7 mL KOH</span>
Answer:
To know this
get the pH
pH= -log(H+)
pH= -log(1x10^-5)
pH= 5.
It is Acidic.
Acidic ranges from 1 - 7
It falls within this range
Answer:
molarity of acid =0.0132 M
Explanation:
We are considering that the unknown acid is monoprotic. Let the acid is HA.
The reaction between NaOH and acid will be:

Thus one mole of acid will react with one mole of base.
The moles of base reacted = molarity of NaOH X volume of NaOH
The volume of NaOH used = Final burette reading - Initial reading
Volume of NaOH used = 22.50-0.55= 21.95 mL
Moles of NaOH = 0.1517X21.95=3.33 mmole
The moles of acid reacted = 3.33 mmole
The molarity of acid will be = 