Question

Calculate either [H3O+] or [OH−] for each of the solutions at 25 °C. Solution A: [OH−]=3.29×10−7 M; [H3O+]= _________M Solution B: [H3O+]=9.63×10−9 M; [OH−]= ____________M Solution C: [H3O+]=6.49×10−4 M; [OH−]= __________M Which of these solutions are basic at 25 °C? Solution A: [OH−]=3.29×10−7 M Solution C: [H3O+]=6.49×10−4 M Solution B: [H3O+]=9.63×10−9 M

Answer 1

Solution A and B are BASIC

`\begin{gathered} solution\text{ A: }[H_3O^+]=3.039*10^(-8)M \\ solution\text{ B: }[OH^-]=1.038*10^(-6)M \\ solution\text{ C: }[OH^-]=1.54*10^(-11)M \end{gathered}`

Explanations:

Using the formula below to determine the concentration of the hydroxonium ion;

`\begin{gathered} [H_3O^+][OH^-]=1*10^(-14) \\ [H_3O^+]=(1*10^(-14))/(3.29*10^(-7)) \\ [H_3O^+]=(1)/(3.29)*10^(-14-(-7)) \\ [H_3O^+]=0.3039*10^{^(-7)} \\ [H_3O^+]=3.039*10^(-8)M \end{gathered}`

Determine the pH

`\begin{gathered} pH=-log[H_3O] \\ pH=-log(3.039*10^(-8)) \\ pH=-(-7.5) \\ pH=7.5 \end{gathered}`

This shows that the solution A is BASIC since the pH is greater than 7

For solution B

`\begin{gathered} [OH^-]=(1*10^(-14))/([H_3O^+]) \\ [OH^-]=(1*10^(-14))/(9.63*10^(-9)) \\ [OH^-]=0.1038*10^(-5)M \\ [OH^-]=1.038*10^(-6)M \end{gathered}`

Determine the pH

`\begin{gathered} pH=-log[H_3O^+] \\ pH=-log(9.63*10^(-9)) \\ pH=-(-8.02) \\ pH=8.02 \end{gathered}`

This shows that the solution B is BASIC since the pH is greater than 7

For the solution C

`\begin{gathered} [OH^-]=(1*10^(-14))/([H_3O^+]) \\ [OH^-]=(1*10^(-14))/(6.49*10^(-4)) \\ [OH^-]=0.1540*10^(-10)M \\ [OH^-]=1.54*10^(-11)M \end{gathered}`

Determine the pH of the solution C

`\begin{gathered} pH=-log[H_3O^+] \\ pH=-log(6.49*10^(-4)) \\ pH=-(-3.19) \\ pH=3.19 \end{gathered}`

Since the pH of the soution C is less than 7, hence the solution C is ACIDIC