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

Dec 28, 2015

$\text{pH} = 9.16$

#### Explanation:

The idea here is that sodium cyanate, $\text{NaOCN}$, will dissociate completely in aqueous solution to form neutral sodium cations, ${\text{Na}}^{+}$, and basic cyanate anions, ${\text{NCO}}^{-}$.

The cyanate anions will react with water to form isocyanic acid, $\text{HNCO}$, and hydroxide anions, ${\text{OH}}^{-}$. 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

$\text{1 cm"^3 = "1 mL" = 10^(-3)"L}$

You will thus have

$\textcolor{b l u e}{c = \frac{n}{V}}$

$c = \text{0.10 moles"/(250 * 10^(-3)"L") = "0.40 M}$

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

${\text{NaOCN"_text((aq]) -> "Na"_text((aq])^(+) + "NCO}}_{\textrm{\left(a q\right]}}^{-}$

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

${\text{ " "NCO"_text((aq])^(-) + "H"_2"O"_text((aq]) " "rightleftharpoons" " "HNCO"_text((aq]) " "+" " "OH}}_{\textrm{\left(a q\right]}}^{-}$

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

$\textcolor{b l u e}{{K}_{W} = {K}_{a} \cdot {K}_{b} \implies {K}_{b} = {K}_{W} / {K}_{a}} \text{ }$, 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=rov=onepage&q&f=false

${K}_{a} = 1.2 \cdot {10}^{- 4}$

This means that you have

${K}_{b} = {10}^{- 14} / \left(1.2 \cdot {10}^{- 4}\right) = 8.33 \cdot {10}^{- 11}$

This will get you

${K}_{b} = \left(\left[{\text{HNCO"] * ["OH"^(-)])/(["NCO}}^{-}\right]\right)$

$8.33 \cdot {10}^{- 11} = \frac{x \cdot x}{0.40 - x} = {x}^{2} / \left(0.40 - x\right)$

Because ${K}_{b}$ is so small, you can use the approximation

$0.40 - x \approx 0.40$

The value of $x$ will thus be

$8.33 \cdot {10}^{- 11} = {x}^{2} / 0.40$

$x = \sqrt{\frac{8.33}{0.40} \cdot {10}^{- 11}} = 1.44 \cdot {10}^{- 5}$

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"^(-)])#

$\text{pOH} = - \log \left(1.44 \cdot {10}^{- 5}\right) = 4.84$

The pH of the solution will be

$\textcolor{b l u e}{\text{pH" + "pOH} = 14}$

$\text{pH} = 14 - 4.84 = \textcolor{g r e e n}{9.16}$

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