Answer
[tex]\begin{gathered} pH\text{ of 4.52 = 3.02}\times10^{-5}\text{moldm}^{-3} \\ pH\text{ of 7.00 = }1\times10^{-7}\text{moldm}^{-3} \end{gathered}[/tex]Explanations:
To calculate the pH of an aqueous solution, you need to know the concentration of the hydronium ion in moles per liter (molarity). The pH is therefore calculated using the formula;
[tex]\begin{gathered} pH=-\log _{10}\lbrack H^{+^{}}(aq)_{}\rbrack \\ -pH=\log _{10}\lbrack H^+(aq)\rbrack \end{gathered}[/tex]Note that the square bracket denotes concentration.
Recall from the law of logarithm;
[tex]\begin{gathered} \text{If log}_ab=x^{} \\ b=a^x \end{gathered}[/tex]Applying this rule to the pH formula above, we will have:
[tex]\lbrack H^+(aq)\rbrack=10^{-pH}[/tex]For the concentration with a pH of 4.52, its concentration will be derived by substituting this pH value into the concentration formula as shown:
[tex]\begin{gathered} \lbrack H^+(aq)\rbrack=10^{-4.52} \\ \lbrack H^+(aq)\rbrack=3.02\times10^{-5}moldm^{-3} \end{gathered}[/tex]Similarly for the concentration of pH of 7.00;
[tex]\begin{gathered} \lbrack H^+(aq)\rbrack=10^{-7} \\ \lbrack H^+(aq)\rbrack=1\times10^{-7}\text{moldm}^{-3} \end{gathered}[/tex]
