Answer:
The correct answer is 0.373 M.
Explanation:
The reaction will be,
H3PO4 (aq) + 3KOH (aq) ⇔ 3H2O (l) + K3PO4 (aq)
Based on the given information, the molarity of the solution KOH is 0.100 M, the volume of the solution given is 112 ml or 0.112 L.
The molarity of the solution with respect to KOH is,
M = Moles/Volume of Solution (liters)
Moles of solute = Molarity * Volume of solution
Moles = 0.100 M * 0.112 L
= 0.0112 moles
From the reaction it is clear that 3 moles of KOH needed 1 mole of H3PO4, thus, 0.0112 moles of KOH will require,
1/3 * 0.0112 = 0.00373 moles of H3PO4
The volume of the H3PO4 solution is 10 ml or 0.01 L.
The molarity of the solution will be,
M = Moles/ Volume of Solution (liters)
M = 0.00373/0.01 = 0.373 M
Thus, the concentration of the solution is 0.373 M.
The concentration of the H3PO4 would be 0.37 M
Going by the balanced equation of the reaction:
[tex]3KOH + H_3PO_4 ---> 3H_2O + K_3PO_4[/tex]
The mole ratio of KOH to H3PO4 is 3:1.
Using the equation: CaVa/CbVb = 1/3
What we are looking for is Ca:
Ca = CbVb/3Va
= 0.1 x 112/3 x 10
= 0.37 M
More or calculations involving neutralization reactions can be found here: https://brainly.com/question/7305038?referrer=searchResults
