a volume of 42.5ml 0.125M KOH is needed to neutalize completely 37.2ml of H3PO4 solution. Find the molar concentation of H3PO4?
normalitybase*volumebase=normality acid*volume acid
normality acid=.125*42.5/37.2
molarity acid= 1/3 normality acid and you have it.
To find the molar concentration of H3PO4, we can use the concept of stoichiometry.
First, let's determine the number of moles of KOH using the given volume and molar concentration:
Moles of KOH = Volume (L) x Molar Concentration (mol/L)
= 42.5 ml x 0.125 mol/L
= 0.0425 L x 0.125 mol/L
= 0.0053125 mol
Since KOH reacts completely with H3PO4 in a 1:1 ratio, the number of moles of H3PO4 in the solution is also 0.0053125 mol.
Next, we calculate the molar concentration of H3PO4 using the moles of H3PO4 and the volume:
Molar Concentration of H3PO4 = Moles of H3PO4 / Volume (L)
= 0.0053125 mol / 0.0372 L
= 0.1429 mol/L (rounded to four decimal places)
Therefore, the molar concentration of H3PO4 is approximately 0.1429 mol/L.
To find the molar concentration of H3PO4, we can use the concept of stoichiometry and the given volume and molarity of KOH.
Let's break down the information given:
Volume of KOH solution (V1) = 42.5 mL
Molarity of KOH solution (M1) = 0.125 M
Volume of H3PO4 solution (V2) = 37.2 mL
Now, we need to use the balanced chemical equation between KOH and H3PO4 to determine the stoichiometry ratio of the two substances.
The balanced chemical equation for the reaction is:
3 KOH + H3PO4 -> K3PO4 + 3 H2O
From the equation, we can see that it takes 3 moles of KOH to react with 1 mole of H3PO4.
Using this stoichiometry ratio, we can set up the following equation:
(Molarity of KOH) x (Volume of KOH) x (Stoichiometry ratio) = (Molarity of H3PO4) x (Volume of H3PO4)
(0.125 M) x (42.5 mL) x (3/1) = (Molarity of H3PO4) x (37.2 mL)
Simplifying the equation, we have:
Molarity of H3PO4 = (0.125 M x 42.5 mL x 3) / 37.2 mL
Molarity of H3PO4 ≈ 0.428 M
Therefore, the molar concentration of H3PO4 is approximately 0.428 M.