Answer:
Here's what I get
Explanation:
3. Molar concentration by formula.
[tex]\begin{array}{rcl}M_{\text{a}}V_{a} & = & M_{\text{b}}V_{b}\\M_{a} \times \text{0.025 00 L} & = & \text{0.3840 mol/L} \times \text{0.034 52 L}\\0.025 00M_{a}\text{ L} & = & \text{0.013 26 mol}\\M_{a}&= &\dfrac{\text{0.013 26 mol}}{\text{0.025 00 L}}\\\\& = &\textbf{0.5302 mol/L}\end{array}[/tex]
(i) Comparison of molar concentrations
The formula gives a calculated value of 0.5302 mol·L⁻¹.
Dimensional analysis gives a calculated value of 0.1767 mol·L⁻¹.
The first value is three times the second.
It is wrong because the formula assumes that the acid supplies just enough moles of H⁺ to neutralize the OH⁻ from the NaOH.
Instead, I mol of H₃PO₄ provides 3 mol of H⁺, so your calculated concentration is three times the true value.
(ii) When is the formula acceptable?
The formula is acceptable only when the molar ratio of acid to base is 1:1.
Examples are
HCl + NaOH ⟶ NaCl + H₂O
H₂SO₄ + Ca(OH)₂ ⟶ CaSO₄ + 2H₂O
H₃PO₄ + Al(OH)₃ ⟶ AlPO₄ + 3H₂O
