the ph of a 1.00^10(-2) M solution of cyanic acid (HOCN) is 2.77 at 25 C. Calculate Ka and pKa for HOCN from this result.
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Hi 1907!
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. So lets plug everything we know into our equation: Ka = [x][x]/(1.00x10^-2). Notice how I put 2 x's on the numerator? 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 remember 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 feel free to ask more questions!
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. So lets plug everything we know into our equation: Ka = [x][x]/(1.00x10^-2). Notice how I put 2 x's on the numerator? 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 remember 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 feel free to ask more questions!