Question

With all the steps please!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+]= 5.01*10^-14M

d) [OH-]= 0.2M

Step-by-step explanation:

The formula of sodium hydroxide is NaOH. In the molecule there are Na+ ions and OH- that dissociates like this:

`\text{ NaOH}\rightarrow\text{ Na}^++\text{ OH}^-`

That means that 1 mole of NaOH dissociates into 1 mole of Na+ ions and 1 mole of OH-.

So, for a 0.2M solution, 0.2 moles of NaOH will dissociate into 0.2 moles of Na+ ions and 0.2 moles of OH-.

d) [OH-]

From the dissociation of NaOH we know that the concentration of OH- is [OH-]=0.2M.

b) pOH

With the concentration of OH-, we can calulate the pOH:

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

So, the pOH is 0.7.

a) pH

Knowing the pOH and the following formula, we can calculate the pH of the solution:

`\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 that we know thw pH of the solution, 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 5.01*10^-14M.