<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
A) Calcium Dihydrogen Phosphate
B) Iron(II) Sulfate
C) Calcium Carbonate
D) Magnesium Oxide
E) Sodium Nitrite
F) Potassium iodide
Complete Question
The complete question is show on the first uploaded image
Answer:
This is shown on the second,third , fourth and fifth image
Explanation:
This is shown on the second,third , fourth and fifth image
Iron III Chloride has a chemical formula of FeCl₃, while ammonium hydroxide has a chemical formula of NH₄OH.
The <em>balanced equation</em> would be:
FeCl₃ (aq) + 3 NH₄OH (aq) → Fe(OH)₃ (s) + 3 NH₄Cl (aq)
The precipitate is Fe(OH)₃ or iron iii hydroxide.
To find the <em>complete ionic equation</em>, dissociate the compounds in aqueous phases into their ionic forms:
Fe³⁺ + Cl⁻ + NH₄⁺ + 3 OH⁻ --> Fe(OH)₃(s) + NH₄⁺ + Cl⁻
To find the <em>net ionic equation</em>, cancel out like ions that appear both in the reactant and product side:
Fe³⁺ + 3 OH⁻ --> Fe(OH)₃
