how many ml of 0.655 M phosphoric acid solution is required to neutralize 15 ml of 1.284 M NaOH solution, write a balanced chemical equation
3NaOH + H3PO4 ==> Na3PO4 + 3H2O
mols NaOH = M x L = ?
Use the coefficients in the balanced equation to convert mols NaOH to mols H3PO4.
Now M H3PO4 = mols H3PO4/L H3PO4. YOu know mols and M, sole for L and convert to mL.
To answer this question, we need to write the balanced chemical equation for the neutralization reaction between phosphoric acid (H₃PO₄) and sodium hydroxide (NaOH):
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
Now, we can use the balanced chemical equation to determine the stoichiometry between phosphoric acid and sodium hydroxide. From the equation, we see that 1 mole of H₃PO₄ reacts with 3 moles of NaOH.
Given:
- Concentration (C₁) of H₃PO₄ solution = 0.655 M
- Volume (V₁) of H₃PO₄ solution = unknown
- Concentration (C₂) of NaOH solution = 1.284 M
- Volume (V₂) of NaOH solution = 15 ml = 0.015 L
Using the equation:
C₁V₁ = C₂V₂
Plug in the given values:
(0.655 M)V₁ = (1.284 M)(0.015 L)
Now, solve for V₁:
V₁ = (1.284 M)(0.015 L) / 0.655 M
Calculating this expression, we find V₁ ≈ 0.0296 L, which is equivalent to 29.6 ml.
So, approximately 29.6 ml of the 0.655 M phosphoric acid solution is required to neutralize 15 ml of the 1.284 M NaOH solution.