Calculate the pH of a 0.025 mol/L solution of sodium nitrite.

4 years ago

Answered By Vincent E

First, figure out if the species we're dealing with is an acid or base. In this case, sodium nitrite does not have a hydrogen, so that's a clue that we are dealing with a base, which is nitrite (NO_{2}^{-}) (as an aside, sodium and other cations in these types of problems act as spectator ions which are not important for solving the problem). Lets write the weak base equation for nitrite ion.

We can then construct an ICE table for this equation. Unfortunately it's not easy to draw one on this platform so I'll just describe how to fill in the table. First, we don't consider water since it's liquid. In the "I" row, we put 0.025 mol/L under NO_{2}^{-} since that is the initial concentration. We put 0 for both OH^{-} and HNO_{2} since those are not present initially. In the "C" row, we put how much each species changes by. We get this from the coefficients in front of the species. We represent the change as 'x' since we don't know how much each species changes by. For NO_{2}^{-}, you would put '-x', and for both OH^{-} and HNO_{2}, we put '+x'. For the "E" row, we put the concentration of each species at equilibrium. In the table, we would simply just add down the column. So for NO_{2}^{-} it would be 0.025-x, and for both OH^{-} and HNO_{2}, it would be just x. Hope that makes sense!

We can then construct our Kb expression. Why Kb and not Kc or Ka? Because we are dealing with a weak base. Therefore,

$Kb=\frac{x\cdot x}{0.025-x}$Kb=x·x0.025−x

We need to solve for x. So we need to know what Kb is. In our data table we are given the Ka for the conjugate acid of NO2-, HNO2. We can use this Ka to find Kb using the following formula:

$Kb=\frac{Kw}{Ka}$Kb=KwKa . Where Ka is the ionization constant of water and is equal to $1.0\times10^{-14}$1.0×10^{−14} . Therefore:

We can check if we can use the approximation method by checking if initial concentration/Kb is greater than 1000. If it is, then we can eliminate the '-x' from the bottom of the Kb expression and make the calculation of x so much easier. Let's check if we can use approximation:

$1.7857\times10^{-11}=\frac{x^2}{0.025}$1.7857×10^{−11}=x^{2}0.025 . We can now easily solve for x using algebra. $x=6.6815\times10^{-7}$x=6.6815×10^{−7} .

In our ICE table, x was the equilibrium concentration for both OH- and HNO2. We can now calculate our pOH.

4 years ago

## Answered By Vincent E

First, figure out if the species we're dealing with is an acid or base. In this case, sodium nitrite does not have a hydrogen, so that's a clue that we are dealing with a base, which is nitrite (NO

_{2}^{-}) (as an aside, sodium and other cations in these types of problems act as spectator ions which are not important for solving the problem). Lets write the weak base equation for nitrite ion.NO

_{2}^{-}_{(aq)}+ H_{2}O_{(l)}-> OH^{-}_{(aq)}+ HNO_{2(aq)}We can then construct an ICE table for this equation. Unfortunately it's not easy to draw one on this platform so I'll just describe how to fill in the table. First, we don't consider water since it's liquid. In the "I" row, we put 0.025 mol/L under NO

_{2}^{-}since that is the initial concentration. We put 0 for both OH^{-}and HNO_{2}since those are not present initially. In the "C" row, we put how much each species changes by. We get this from the coefficients in front of the species. We represent the change as 'x' since we don't know how much each species changes by. For NO_{2}^{-}, you would put '-x', and for both OH^{-}and HNO_{2}, we put '+x'. For the "E" row, we put the concentration of each species at equilibrium. In the table, we would simply just add down the column. So for NO_{2}^{-}it would be 0.025-x, and for both OH^{-}and HNO_{2}, it would be just x. Hope that makes sense!We can then construct our Kb expression. Why Kb and not Kc or Ka? Because we are dealing with a weak base. Therefore,

$Kb=\frac{x\cdot x}{0.025-x}$Kb=x·x0.025−x

We need to solve for x. So we need to know what Kb is. In our data table we are given the Ka for the conjugate acid of NO2-, HNO2. We can use this Ka to find Kb using the following formula:

$Kb=\frac{Kw}{Ka}$Kb=KwKa . Where Ka is the ionization constant of water and is equal to $1.0\times10^{-14}$1.0×10

^{−14}. Therefore:$Kb=\frac{1.0\times10^{-14}}{5.6\times10^{-4}}=1.7857\times10^{-11}$Kb=1.0×10

^{−14}5.6×10^{−4}=1.7857×10^{−11}.So we have,

$Kb=\frac{x^2}{0.025-x}=1.7857\times10^{-11}$Kb=x

^{2}0.025−x =1.7857×10^{−11}We can check if we can use the approximation method by checking if initial concentration/Kb is greater than 1000. If it is, then we can eliminate the '-x' from the bottom of the Kb expression and make the calculation of x so much easier. Let's check if we can use approximation:

$\frac{0.025}{1.7857\times10^{-11}}=1400000000>1000$0.0251.7857×10

^{−11}=1400000000>1000 . Therefore:$1.7857\times10^{-11}=\frac{x^2}{0.025}$1.7857×10

^{−11}=x^{2}0.025 . We can now easily solve for x using algebra. $x=6.6815\times10^{-7}$x=6.6815×10^{−7}.In our ICE table, x was the equilibrium concentration for both OH- and HNO2. We can now calculate our pOH.

$pOH=-log\left[OH^-\right]=-log\left(6.6815\times10^{-7}\right)=6.175$pOH=−log[OH

^{−}]=−log(6.6815×10^{−7})=6.175Finally, we can calculate pH:

$pH=14-pOH=14-6.175=7.82$pH=14−pOH=14−6.175=7.82

This pH makes sense because we know we had a weak base solution. This was a difficult question! If you need more help please reach out.

4 years ago

## Answered By Vincent E

Apologies, in the above answer, where it says Ka is the ionization constant of water. It should say Kw there instead of Ka.

4 years ago

## Answered By Vincent E

You can find Ka using the data table. Find HNO2 in the table, and it's the very last column.