A solution is usualy a diluted liquid that cleans for example bleach solution.
Answer:

Explanation:
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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:

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Answer:
10.28 mol
Explanation:
S + 2O = SO2
(atm x L) ÷ (0.0821 x K)
(3.45 x 45.6) ÷ (0.0821 x 373)
=5.13726
Then round it to significant figures
=5.14
5.14 mol SO2 x (2 mol O ÷ 1 mol SO2)
=10.28 mol O
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