In a 1.00 x10^-2 M solution of cyanic acid at 25 C, the pH is 2.77. Calculate the Ka and pKa for cyanic acid.
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Hi Gregory!
This problem requires you to use acid equilibrium. So first off, we need to write the acid dissociation of HOCN which is HOCN ---> H+ + OCN-. This is important because this tells us the stoichiometric ratio between each of the substances and they all appear to be in 1:1 ratios, which makes everything so much simpler :). Now for any equilibrium reaction, it is always Keq = concentrations of products/ concentrations of reactants. For acids, we leave everything the same except that Keq is now Ka.
Plug everything we know into our equation: Ka = [x][x]/(1.00x10^-2). I put 2 x's on the numerator because I don't know the concentration of H+ or OCN- at the moment so I used x to represent my unknown (also I can assume that there are equal molar concentrations of H+ and OCN- because in the dissociation equation, everything is in the same ratio). Now pH = -log[H+}, so I can use this equation to solve for the concentration of H+ ions or my x. 2.77 = -log[x], so x equals about 1.70x10^-3M of H+.
Now that we know our x, plug it back into the equilibrium equation and solve for Ka: Ka = [1.70x10^-3][1.70x10^-3]/(1.00x10^-2). So Ka is about 2.88 x 10^-4. We have our first answer but now we want the value of pKa. pKa = -log[Ka}, so now we just plug in our value of Ka and solve for pKa! pKa = -log[2.88x10^-4] so about 3.52.
I hope this helped and good luck with chem :)
This problem requires you to use acid equilibrium. So first off, we need to write the acid dissociation of HOCN which is HOCN ---> H+ + OCN-. This is important because this tells us the stoichiometric ratio between each of the substances and they all appear to be in 1:1 ratios, which makes everything so much simpler :). Now for any equilibrium reaction, it is always Keq = concentrations of products/ concentrations of reactants. For acids, we leave everything the same except that Keq is now Ka.
Plug everything we know into our equation: Ka = [x][x]/(1.00x10^-2). I put 2 x's on the numerator because I don't know the concentration of H+ or OCN- at the moment so I used x to represent my unknown (also I can assume that there are equal molar concentrations of H+ and OCN- because in the dissociation equation, everything is in the same ratio). Now pH = -log[H+}, so I can use this equation to solve for the concentration of H+ ions or my x. 2.77 = -log[x], so x equals about 1.70x10^-3M of H+.
Now that we know our x, plug it back into the equilibrium equation and solve for Ka: Ka = [1.70x10^-3][1.70x10^-3]/(1.00x10^-2). So Ka is about 2.88 x 10^-4. We have our first answer but now we want the value of pKa. pKa = -log[Ka}, so now we just plug in our value of Ka and solve for pKa! pKa = -log[2.88x10^-4] so about 3.52.
I hope this helped and good luck with chem :)