Answer:

Explanation:
Hello there!
In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

And the concentrations are:
![[acid]=0.000855mol/0.025L=0.0342M](https://tex.z-dn.net/?f=%5Bacid%5D%3D0.000855mol%2F0.025L%3D0.0342M)
![[base]=0.000781mol/0.025L=0.0312M](https://tex.z-dn.net/?f=%5Bbase%5D%3D0.000781mol%2F0.025L%3D0.0312M)
Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

Best regards!
Find one mole
8 C = 8 * 12 = 96
9 H = 1 * 9 = 9
4 O = 4 *16 = 64
Total = 169
1 mol = 169 grams.
0.432 mol = x
1/0.432 = 169/x
x = 0.432 * 169
x = 73.0 grams
Manipulated variable: the types of soil
Responding variable: how much t the plants grow
the numbers are going to be small so like a power but its at the bottom
NH3, H2O2, NHO2