Answer:
[H₃O⁺] = 1.4 × 10⁻⁹ M.
Explanation:
NH₄Cl is a salt that dissolves well in water. The 2.5 M NH₄Cl will give an initial NH₄⁺ concentration of 2.5 M.
NH₃ is a weak base. It combines with water to produce NH₄⁺ and OH⁻. The opposite process can also take place. NH₄⁺ combines with OH⁻ to produce NH₃ and H₂O. The final H₃O⁺ concentration can be found from the OH⁻ concentration. What will be the final OH⁻ concentration?
Let the increase in OH⁻ concentration be x. The initial OH⁻ concentration at room temperature is 10⁻⁷ M.
Construct a RICE table for the equilibrium between NH₃ and NH₄⁺:
.
The 
value for ammonia is small. The value of x will be so small that at equilibrium, 
and 
.
![\displaystyle \text{K}_b = \frac{[{\text{NH}_4}^{+}]\cdot [{\text{OH}}^{-}]}{[\text{NH}_3]} \approx \frac{2.5\;(x + 1.0\times 10^{-7})}{1.0}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BK%7D_b%20%3D%20%5Cfrac%7B%5B%7B%5Ctext%7BNH%7D_4%7D%5E%7B%2B%7D%5D%5Ccdot%20%5B%7B%5Ctext%7BOH%7D%7D%5E%7B-%7D%5D%7D%7B%5B%5Ctext%7BNH%7D_3%5D%7D%20%5Capprox%20%5Cfrac%7B2.5%5C%3B%28x%20%2B%201.0%5Ctimes%2010%5E%7B-7%7D%29%7D%7B1.0%7D)
.
.
![\displaystyle [\text{OH}^{-}] = x+1.0\times 10^{-7} = 1.8\times 10^{-5} /\left(\frac{2.5}{1.0}\right) = 7.2\times 10^{-6}\;\text{mol}\cdot\text{L}^{-}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20x%2B1.0%5Ctimes%2010%5E%7B-7%7D%20%3D%201.8%5Ctimes%2010%5E%7B-5%7D%20%2F%5Cleft%28%5Cfrac%7B2.5%7D%7B1.0%7D%5Cright%29%20%3D%207.2%5Ctimes%2010%5E%7B-6%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7BL%7D%5E%7B-%7D)
.
Again, 
at room temperature.
![\displaystyle [\text{H}_3\text{O}^{+}] = \frac{\text{K}_w}{[\text{OH}^{-}]}=\frac{1.0\times 10^{-14}}{7.2\times 10^{-6}} = 1.4\times 10^{-9} \;\text{mol}\cdot\text{L}^{-1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BH%7D_3%5Ctext%7BO%7D%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B%5Ctext%7BK%7D_w%7D%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%7D%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-14%7D%7D%7B7.2%5Ctimes%2010%5E%7B-6%7D%7D%20%3D%201.4%5Ctimes%2010%5E%7B-9%7D%20%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7BL%7D%5E%7B-1%7D)
Answer/Explanation:
Methanol has a molecular weight (32.04 g/mol), low-boiling point and because of its low boiling point, methanol readily evaporates at room temperature.
Under these specified non-standard conditions, the partial pressure of methanol is lower than its vapor pressure and this explains the reason for the spontaneous evaporation exhibited by methanol.
Answer:
Explanation:
Given parameters;
pH = 8.74
pH = 11.38
pH = 2.81
Unknown:
concentration of hydrogen ion and hydroxyl ion for each solution = ?
Solution
The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.
It is graduated from 1 to 14
pH = -log[H₃O⁺]
pOH = -log[OH⁻]
pH + pOH = 14
Now let us solve;
pH = 8.74
since pH = -log[H₃O⁺]
8.74 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.82 x 10⁻⁹mol dm³
pH + pOH = 14
pOH = 14 - 8.74
pOH = 5.26
pOH = -log[OH⁻]
5.26 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] = 5.5 x 10⁻⁶mol dm³
2. pH = 11.38
since pH = -log[H₃O⁺]
11.38 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 4.17 x 10⁻¹² mol dm³
pH + pOH = 14
pOH = 14 - 11.38
pOH = 2.62
pOH = -log[OH⁻]
2.62 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =2.4 x 10⁻³mol dm³
3. pH = 2.81
since pH = -log[H₃O⁺]
2.81 = -log[H₃O⁺]
[H₃O⁺] = 10⁻
[H₃O⁺] = 1.55 x 10⁻³ mol dm³
pH + pOH = 14
pOH = 14 - 2.81
pOH = 11.19
pOH = -log[OH⁻]
11.19 = -log[OH⁻]
[OH⁻] = 10
[OH⁻] =6.46 x 10⁻¹²mol dm³
