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The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionizationin the solution. |
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Answer» Solution :Calculation of pH `[H^+]` on base of pH : `pH=-log [H^+]=-log (2.34)=-log bar3.76` `=4.157xx10^(-3)` `therefore [H^+] = 10^(-2.34) = 4.57xx10^(-3)` M Calculation of ionization constant `(K_a)`: SUPPOSE ionized HCNO=X M At equili. [HCNO] = (0.1 -x) `approx` 0.1 and `[CNO^-]=[H_3O^+]=xM` `{:(,HCNO_((aq))+ ,H_2O_((l)) HARR, CNO_((aq))^(-)+ , H_3O_((aq))^(+)),("Initially M:",0.1,-,0.0,0.0),("At equilibrium M",(0.1-x),-,x M, x M),(,approx 0.1 M, ,=4.57xx10^(-3)M, =4.57xx10^(-3)M):}` `K_a=([CNO^-][H_3O^+])/([HCNO])` `=((4.57xx10^(-3))(4.57xx10^(-3)))/0.1= 2.088xx10^(-4)` Calculation of degree of ionization of HCNO : Suppose degree of ionization = `alpha` So, `(C xx alpha) =0.1 alpha `, HCNO is ionized , `[H^+]=[CNO^-]`= (ionized HCNO) `therefore [H^+] = 0.1 alpha` According to above calculation `[H^+]=4.57xx10^(-3)` M `therefore 0.1 alpha = 4.57xx10^(-3)` `therefore alpha = 4.57xx10^(-2)` =0.0457 Thus degree of ionization of HCNO = 0.0457 `=4.57xx10^(-2)` |
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