Question

DUE TODAY EXTREMELY IMPORTANT PLEASE PLEASE PLEASE HELP ME ASAPPPPPPPP CHEMISTRY!!!!!!!!!!!!!!!!!!! I WOULD APPRECIATE IT ALOT THANK YOU IN ADVANCE!!!!!!!!!!!!!!!!!!!

Answer 1

Here's what I get

Step-by-step explanation:

3. Molar concentration by formula.

`\begin{array}{rcl}M_{\text{a}}V_(a) & = & M_{\text{b}}V_(b)\\M_(a) * \text{0.025 00 L} & = & \text{0.3840 mol/L} * \text{0.034 52 L}\\0.025 00M_(a)\text{ L} & = & \text{0.013 26 mol}\\M_(a)&= &\frac{\text{0.013 26 mol}}{\text{0.025 00 L}}\\\\& = &\textbf{0.5302 mol/L}\end{array}`

(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