Answer:
pH = 6.82
Explanation:
To solve this problem we can use the<em> Henderson-Hasselbach equation</em>:
- pH = pKa + log
![\frac{[NaOCl]}{[HOCl]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNaOCl%5D%7D%7B%5BHOCl%5D%7D)
We're given all the required data to <u>calculate the original pH of the buffer before 0.341 mol of HCl are added</u>:
- pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54
- [HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M
- pH = 7.54 + log

By adding HCl, w<em>e simultaneously </em><u><em>increase the number of HOCl</em></u><em> and </em><u><em>decrease NaOCl</em></u>:
- pH = 7.54 + log
![\frac{[NaOCl-HCl]}{[HOCl+HCl]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNaOCl-HCl%5D%7D%7B%5BHOCl%2BHCl%5D%7D)
- pH = 7.54 + log

Answer:
7.21 grams is the mass of methane
Explanation:
We may use the Ideal Gases Equation to solve this:
P. V = n. R. T
Let's determine the moles of Ar
18 g . 1 mol/ 39.9 g = 0.451 mol
In both situations, volume, temperature and pressure are the same so the moles of methane will also be the same as Argon's.
Let's convert the moles to mass of CH4.
0.451 mol . 16g/1mol = 7.21 grams
D. The final substance in Beaker A is a mixture and in Beaker B is a pure substance.