Question

With all the steps please for a test!!For a 0.2 M aqueous solution of sodium hydroxide, calculate the following:a) pHb) pOHc) [H+]d) [OH-]

Answer 1

a) pH= 13.3

b) pOH= 0.7

c) [H+]= 5.01x10^-14M

d) [OH-]= 0.2M

Step-by-step explanation:

The formula of sodium hydroxide is NaOH. From the dissociation of 1 mole of NaOH we obtain 1 mole of Na+ ion and 1 mole of OH- ion:

`NaOH\rightarrow Na^++OH^-`

d) [OH-]

From the dissociation of 0.2M NaOH we will obtain a concentration of 0.2M Na+ and 0.2M OH-.

So, the concentration of OH- is 0.2M.

b) pOH

With the concentration of OH- and the formula of pOH we can calculate the pOH of the solution:

`\begin{gathered} pOH=-log\lbrack OH^-\rbrack \\ pOH=-log(0.2) \\ pOH=0.7 \end{gathered}`

The pOH of the solution is 0.7.

a) pH

We can calculate the pH of the solution using the following formula, and replacing the value of pOH:

`\begin{gathered} pH+pOH=14 \\ pH+0.7=14 \\ pH=14-0.7 \\ pH=13.3 \end{gathered}`

The pH of the solution is 13.3.

c) [H+]

Now with the pH formula we can calculate the concentration of H+:

`\begin{gathered} pH=-log\lbrack H^+\rbrack \\ 13.3=-log\lbrack H^+\rbrack \\ 10^((-13.3))=\lbrack H^+\rbrack \\ 5.01*10^(-14)=\lbrack H^+\rbrack \end{gathered}`

So, the concentration of H+ is 5.01x10^-14M.