PH = 0.1289<span> for </span>1.50<span> M solution of weak acid with Ka value of </span><span>.73</span>
Answer: The concentration of hydrogen ions for this solution is 
.
Explanation:
Given: pOH = 11.30
The relation between pH and pOH is as follows.
pH + pOH = 14
pH + 11.30 = 14
pH = 14 - 11.30
= 2.7
Also, pH is the negative logarithm of concentration of hydrogen ions.
![pH = - log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5BH%5E%7B%2B%7D%5D)
Substitute the values into above formula as follows.
![pH = -log [H^{+}]\\2.7 = -log [H^{+}]\\conc. of H^{+} = 1.99 \times 10^{-3}](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5C2.7%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5Cconc.%20of%20H%5E%7B%2B%7D%20%3D%201.99%20%5Ctimes%2010%5E%7B-3%7D)
Thus, we can conclude that the concentration of hydrogen ions for this solution is .
Answer:
384.2 K
Explanation:
First we convert 27 °C to K:
- 27 °C + 273.16 = 300.16 K
With the absolute temperature we can use <em>Charles' law </em>to solve this problem. This law states that at constant pressure:
Where in this case:
We input the data:
300.16 K * 1600 m³ = T₂ * 1250 m³
And solve for T₂:
T₂ = 384.2 K
Prepare a 1% copper sulfate solution. To make this solution, weigh 1 gram of copper sulfate (CuSO4 ·5H2O), dissolve in a small amount of distilled water in a 100 ml volumetric flask and bring to volume. Label this as 1% copper sulfate solution.
That would be either, nuclear power or wind power, depending on what the teacher is looking for...
