Acids and Bases Weak Acids and Bases

The pH of 1.00 x 10^-2 M solution of cyanic acid (HOCN) is 2.77 at 25 degrees Celsius. Calculate Ka for HOCN from this result.

To calculate the acid-ionization constant for cyanic acid, HOCN, the balanced chemical equation showing the ionization of cyanic acid is needed so we can then write the acid-ionization expression to eventually solve for Ka, the acid-ionization constant.

HOCN (aq) <—> H+ (aq) + OCN- (aq)

Ka = [H+][OCN-] / [HOCN]

To begin to piece together the equilibrium values needed for the Ka expression, an ICE table is used to map out the information given. An initial solution of 1.00 E-2 M HOCN is given:

HOCN<–>H+OCN-
initial1.00E-2 M0 M0 M
change-x+x+x
equilibrium

The pH of acid-base reactions is measured after a solution has reached equilibrium so we can use the pH value of 2.77 given to calculate the equilibrium concentration of hydrogen ions:

10^-pH = [H+] therefore, 10^-2.77 = [H+] = 1.698 E-3 M

HOCN<–>H+OCN-
initial1.00E-2 M0 M0 M
change-x+x+x
equilibrium1.698E-3 M

The initial concentration of H+ is 0 M and since we now know the equilibrium concentration of H+, we can solve for the change variable, x, needed to fill in the rest of the values in the ICE table:

0 + x = 10^-2.77 therefore, x = 10^-2.77 = 1.698E-3 M

HOCN<–>H+OCN-
initial1.00E-2 M0 M0 M
change– 1.698E-3 M + 1.698E-3 M + 1.698E-3 M
equilibrium8.30E-3 M 1.698E-3 M 1.698E-3 M

Now, that the equilibrium values are known for cyanate, OCN-, hydrogen ion, H+, and cyanic acid, HOCN, the acid-ionization equilibrium expression can be used to calculate the value of Ka, the acid-ionization constant.

HOCN (aq) <—> H+ (aq) + OCN- (aq)

[H+] = [OCN-] = 1.698E-3 M

[HOCN] = 8.30E-3 M

Ka = [H+][OCN-]/ [HOCN]

Ka= ( 1.698E-3 M)^2/ (8.30E-3 M)

HOCN Ka = 3.47E-4

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