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What is the pH of a solution of [tex]$H_2SO_4$[/tex] that has [tex]\left[ H_3O^+ \right]=5.45 \times 10^{-5} M[/tex]?

Round to the nearest hundredth.

[tex]\boxed{}

Sagot :

To determine the pH of a solution with a given hydronium ion concentration, we use the formula for pH, which is defined as:

[tex]\[ \text{pH} = -\log_{10} [\text{H}_3\text{O}^+] \][/tex]

Given the concentration of hydronium ions [tex]\(\left[ \text{H}_3\text{O}^+ \right]\)[/tex] is [tex]\(5.45 \times 10^{-5}\)[/tex] M, we follow these steps to calculate the pH:

1. Identify the hydronium ion concentration:
[tex]\[ [\text{H}_3\text{O}^+] = 5.45 \times 10^{-5} \text{ M} \][/tex]

2. Calculate the pH:
[tex]\[ \text{pH} = -\log_{10} (5.45 \times 10^{-5}) \][/tex]
After applying the logarithm function, we obtain a pH value of approximately 4.2636.

3. Round the pH to the nearest hundredth:
[tex]\[ \text{pH} \approx 4.26 \][/tex]

Therefore, the pH of the solution is 4.26 when rounded to the nearest hundredth.