Answer:
1. H3PO4 + 3NaOH —> Na3PO4 + 3H2O
2. 38.5mL
Explanation:
1. We'll begin by writing a balanced equation for the reaction. This is illustrated below:
H3PO4 + 3NaOH —> Na3PO4 + 3H2O
2. H3PO4 + 3NaOH —> Na3PO4 + 3H2O
From the equation above, the following data were obtained:
nA (mole of the acid) = 1
nB (mole of the base) = 3
Data obtained from the question include:
Vb (volume of base) =?
Mb (Molarity of base) = 0.120 M
Va (volume of acid) = 35.0 mL
Ma (Molarity of acid) = 0.0440 M
Using the formula MaVa/MbVb = nA/nB, the volume of the base (i.e NaOH) can be obtained as follow:
MaVa/MbVb = nA/nB
0.0440 x 35/ 0.120 x Vb = 1/3
Cross multiply to express in linear form as shown below:
0.120 x Vb = 0.0440 x 35 x 3
Divide both side by 0.120
Vb = (0.0440 x 35 x 3) /0.120
Vb = 38.5mL
Therefore, 38.5mL of 0.120 M NaOH is needed for the complete neutralization.
Answer:
We need 38.5 mL of NaOH to neutralize the H3PO4 solution
Explanation:
Step 1: Data given
Molarity of NaOH = 0.120 M
Volume of H3PO4 = 35.0 mL = 0.035 L
Molarity of H3PO4 = 0.0440 M
Step 2: The balanced equation
H3PO4 + 3NaOH —> Na3PO4 + 3H2O
Step 3: Calculate the volume of NaOH
b*Ca*Va = a *Cb*Vb
⇒with b = the coefficient of NaOH = 3
⇒with Ca = the concentration of H3PO4 = 0.0440 M
⇒with Va = the volume of H3PO4 = 35.0 mL = 0.0350 L
⇒with a = the coefficient of H3PO4 = 1
⇒with Cb = the concentration of NaOH = 0.120 M
⇒with Vb = the volume of NaOH = TO BE DETERMINED
3*0.0440 * 0.0350 = 0.120 * Vb
Vb = 0.0385 L = 38.5 mL
We need 38.5 mL of NaOH to neutralize the H3PO4 solution
