The correct answer is D. magnetic bond
Answer : The pH of the solution is, 9.63
Explanation : Given,
The dissociation constant for HCN = 
First we have to calculate the moles of HCN and NaCN.

and,

The balanced chemical reaction is:

Initial moles 0.1116 0.0461 0.08978
At eqm. (0.1116-0.0461) 0 (0.08978+0.0461)
0.0655 0.1359
Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:


Therefore, the pH of the solution is, 9.63
If it is assumed that there are only two isotopes then the percent abundance needs to add up to 100%
100-35= 65%
The second isotope will have a 65% abundance.
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To determine the empirical formula and the molecular formula of the compound, we assume a basis of the compound of 100 g. We do as follows:
Mass Moles
K 52.10 52.10/39.10 = 1.33 1.33/1.32 ≈ 1
C 15.8 15.8/12 = 1.32 1.32/1.32 ≈ 1
O 32.1 32.1 / 16 = 2.01 2.01/1.32 ≈ 1.5
The empirical formula would most likely be KCO.
The molecular formula would be K2C2O3.