0.10 moles of sodium cyanate dissolved in 250cm^3 distilled water ,calculate pH?

The idea here is that sodium cyanate, #"NaOCN"#, will dissociate completely in aqueous solution to form neutral sodium cations, #"Na"^(+)#, and basic cyanate anions, #"NCO"^(-)#.

The cyanate anions will react with water to form isocyanic acid, #"HNCO"#, and hydroxide anions, #"OH"^(-)#. This tells you that you can expect the pH of the resulting solution to be higher than #7#.

Calculate the molarity of the sodium cyanate solution by using the given number of moles and volume. Keep in mind that you need to use liters of solution and that

#"1 cm"^3 = "1 mL" = 10^(-3)"L"#

You will thus have

#color(blue)(c = n/V)#

#c = "0.10 moles"/(250 * 10^(-3)"L") = "0.40 M"#

Sodium cyanate dissociates in a #1:1# mole ratio with the cyanate anions

#"NaOCN"_text((aq]) -> "Na"_text((aq])^(+) + "NCO"_text((aq])^(-)#

This means that the concentration of the cyanate anions will be equal to that of the dissolved salt.

#["NCO"^(-)] = "0.40 M"#

Use an ICE table to help you find the equilibrium concentration of hydroxide anions

#" " "NCO"_text((aq])^(-) + "H"_2"O"_text((aq]) " "rightleftharpoons" " "HNCO"_text((aq]) " "+" " "OH"_text((aq])^(-)#

#color(purple)("I")" " " "0.40" " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " "0#
#color(purple)("C")" "(-x)" " " " " " " " " " " " " " " " " "(+x)" " " " " " " "(+x)#
#color(purple)("E")" "0.40-x" " " " " " " " " " " " " " " " " "x" " " " " " " " " " " "x#

By definition, the base dissociation constant, #K_b#, is equal to

#color(blue)(K_W = K_a * K_b implies K_b = K_W/K_a)" "#, where

#K_W# - the ion product constant for water, equal to #10^(-14)# at room temperature
#K_a# - the acid dissociation constant of the conjugate acid

Since you didn't provide a value for #K_b#, I will use the value of #K_a# to find it. I was able to track it down here - page 290, problem 124

https://books.google.ro/books?id=hsuV9JTGaP8C&printsec=frontcover&hl=ro#v=onepage&q&f=false

#K_a = 1.2 * 10^(-4)#

This means that you have

#K_b = 10^(-14)/(1.2 * 10^(-4)) = 8.33 * 10^(-11)#

This will get you

#K_b = (["HNCO"] * ["OH"^(-)])/(["NCO"^(-)])#

#8.33 * 10^(-11) = (x * x)/(0.40 -x ) = x^2/(0.40 -x )#

Because #K_b# is so small, you can use the approximation

#0.40 -x ~~ 0.40#

The value of #x# will thus be

#8.33 * 10^(-11) = x^2/0.40#

#x = sqrt(8.33/0.40 * 10^(-11)) = 1.44 * 10^(-5)#

Since #x# represents the equilibrium concentration of hydroxide anions, you will have

#["OH"^(-)] = 1.44 * 10^(-5)"M"#

THe pOH of the solution will be

#"pOH" = - log(["OH"^(-)])#

#"pOH" = - log(1.44 * 10^(-5)) = 4.84#

The pH of the solution will be

#color(blue)("pH" + "pOH" = 14)#

#"pH" = 14 - 4.84 = color(green)(9.16)#

As predicted, the pH of the solution is higher than #7#.