Unfortunately the data provided doesn't include the DENSITY of the ammonium chloride solution and molarity is defined as moles per volume. So without the density, the calculation of the molarity is impossible. But fortunately, there are tables available that do provide the required density and for a 20% solution by weight, the density of the solution is 1.057 g/ml.
So 1 liter of solution will mass 1057 grams and the mass of ammonium chloride will be 0.2 * 1057 g = 211.4 g. The number of moles will then be 211.4 g / 53.5 g/mol = 3.951401869 mol. Rounding to 3 significant digits gives a molarity of 3.95.
Now assuming that your teacher wants you to assume that the solution masses 1.00 g/ml, then the mass of ammonium chloride will only be 200g, and that is only (200/53.5) = 3.74 moles.
So in conclusion, the expected answer is 3.74 M, although the correct answer using missing information is 3.95 M.
Answer:
<em>C</em> NaOHsln = 0.125 M
Explanation:
Molarity (M) [=] mol/L
∴ moles NaOH = (0.125 L)(0.15 mol/L) = 0.01875 mol
∴ Vsln = 150 mL = 0.150 L
New Molarity:
⇒ <em>C</em> NaOH = 0.01875 mol/0.150 L
⇒ <em>C</em> NaOH = 0.125 M
When frequency increases, wavelength increases
Answer:
Explanation:
<em>Percent ionization</em> is the percent of the original acid that has ionized:
- %, ionization = (molar concentration of hydrogen ions at equilibrium / molar concentration of original acid) × 100
<u><em>Part A:</em></u>
<u>1) Data:</u>
- Ka: 6.7 × 10 ⁻⁷
- [HA] = 0.10 M
- %, ionization = ?
<u>2) Equilibrium equation:</u>
<u>3) ICE (initial, change, equilbirium) table </u>
Concentrations
HA H⁺ A⁻
Initial 0.10 0 0
Change - x + x + x
Equilibrium 0.10 - x x x
- Equation: Ka = [H⁺] [A⁻] / [HA] =
6.7 × 10 ⁻⁷ = x² / (0.10 - x)
<u>4) Solve the equation:</u>
Since Ka << 1, you can assume x << 0.10 and 0.10 - x ≈ 0.10
- 6.7 × 10 ⁻⁷ ≈ x² / 0.10 ⇒ x² ≈ 6.7 × 10⁻⁸ ⇒ x ≈ 2.588 × 10⁻⁴
- % ionization ≈ (2.588 × 10⁻⁴ M / 0.1 M) × 100 ≈ 0.2588 % ≈ 0.26% (two significant figures)
<u><em>Part B:</em></u>
<u>1) Data:</u>
- Ka: 6.7 × 10 ⁻⁷
- [HA] = 0.010 M
- %, ionization = ?
<u>2) Equilibrium equation:</u>
<u>3) ICE table:</u>
Concentrations
HA H⁺ A⁻
Initial 0.010 0 0
Change - x + x + x
Equilibrium 0.010 - x x x
- Equation: Ka = [H⁺] [A⁻] / [HA] =
6.7 × 10 ⁻⁷ = x² / (0.010 - x)
<u>4) Solve the equation</u>:
Since Ka << 1, you can assume x << 0.010 and 0.010 - x ≈ 0.010
- 6.7 × 10 ⁻⁷ ≈ x² / 0.010 ⇒ x² ≈ 6.7 × 10⁻⁹ ⇒ x ≈ 8.185 × 10⁻5
- % ionization ≈ (8.185 × 10⁻⁵ M / 0.010 M) × 100 ≈ 0.8185 % ≈ 0.82% (two significant figures)
