Q3- A solution of sodium cyanate- NaCNO(aq)- is prepared at a concentr.docx

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Q3. A solution of sodium cyanate, NaCNO(aq), is prepared at a concentration of 0.20 M. Calculate the equilibrium concentrations of OH (aq), HCNO (aq), CNO (aq), and H30* (aq) and the pH of the solution at 25Â°C Solution NaCNO(aq) -----------------> Na^+ (aq) + CNO^- (aq) 0.2MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.2M CNO^- (aq) + H2O(l) -----------> HCNO(aq) + OH^- (aq) IÂ Â Â Â Â Â Â Â 0.2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0 CÂ Â Â Â Â Â Â Â - xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x EÂ Â Â Â Â Â Â Â 0.2-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x KbÂ Â Â = Kw/Ka = 1*10^-14/3.5*10^-4Â Â Â = 2.86*10^-11 KbÂ Â Â Â Â =Â Â Â [HCNO][OH^-]/[CNO^-] 2.86*10^-11Â Â Â = x*x/0.2-x 2.86*10^-11*(0.2-x)Â Â = x^2 x Â Â = 2.4*10^-6 [CNO^-]Â Â = xÂ Â = 2.4*10^-6M [OH^-]Â Â = xÂ Â = 2.4*10^-6M [H3O^+]Â Â Â = Kw/[OH^-] = 1*10^-14/2.4*10^-6Â Â = 4.16*10^-9M PHÂ Â Â = -log[H3O^+] = -log4.16*10^-9 = 8.38 .

Educação

[CNO^-]Â Â = xÂ Â = 2.4*10^-6M
[OH^-]Â Â = xÂ Â = 2.4*10^-6M
[H3O^+]Â Â Â = Kw/[OH^-]
= 1*10^-14/2.4*10^-6Â Â = 4.16*10^-9M
PHÂ Â Â = -log[H3O^+]
= -log4.16*10^-9
= 8.38

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