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NaOH(s) was added to 1.0 L of HClO4(aq) 0.43 M. Calculate [H3O+] in the solution after the addition of 0.10 mol of NaOH(s).Calculate [H3O+] in the solution after the addition of 0.81 mol of NaOH(s).

Sagot :

Answer:

a) 0.33 M

b) 2.63x10^-14 M

Explanation:

1st) It is necessary to write and balance the chemical reaction:

[tex]HClO_4+NaOH\rightarrow NaClO_4+H_2O[/tex]

From the balanced reaction we can see that 1 mole of HClO4 reacts with 1 mole of NaOH.

2nd) Since the relation between the acid (HClO4) and the base (NaOH) is 1:1, the 0.10 moles of NaOH will react with 0.10 moles of acid:

[tex]\begin{gathered} RemainingSolution=0.43moles\text{ - }0.10moles \\ RemainingSolution=0.33moles \end{gathered}[/tex]

So, in the solution 0.33 moles of HClO4 will remain unreacted, and the concentration of H3O+ will be 0.33M.

3rd) In this case we can proceed the same as in the first part:

[tex]\begin{gathered} RemainingSolution=0.43moles-0.81moles\text{ } \\ RemainingSolution=-0.38moles \end{gathered}[/tex]

In this case, the addition of 0.81 moles of NaOH neutralizes all the acid in the solution and 0.38 moles of the base will remain unreacted.

So, the concentration of OH- will be 0.38M, and it is necessary to calculate the pOH of the solution, because when OH- is left over, the solution will be basic:

[tex]\begin{gathered} pOH=-log\lbrack OH^-\rbrack \\ pOH=-log(0.38) \\ pOH=0.42 \end{gathered}[/tex]

The pOH of the solution is 0.42.

4th) Now with the pOH we can calculate the pH of the solution, with the following formula:

[tex]\begin{gathered} pH+pOH=14 \\ pH+0.42=14 \\ pH=14-0.42 \\ pH=13.58 \end{gathered}[/tex]

The pH of the solution is 13.58.

5th) Finally, with the value of pH and the pH formula, we can calculate [H3O+]:

[tex]\begin{gathered} pH=-log\lbrack H3O^+\rbrack \\ 13.58=-log\lbrack H3O^+\rbrack \\ -13.58=log\lbrack H3O^+\rbrack \\ 10^{(-13.58)}=\lbrack H3O^+\rbrack \\ 2.63*10^{-14}M=\lbrack H3O^+\rbrack \end{gathered}[/tex]

So, the concentration of H3O+ is 2.63x10^-14M.

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