Question

the ph of a 0.0054 m solution of hno₃ is

Explanation:

Step1: Determine the concentration of \(H^+\) ions

Since \(HNO_3\) is a strong acid, it completely dissociates in water as \(HNO_3
ightarrow H^++NO_3^-\). So, \([H^+]=0.00544\space M\)

Step2: Calculate the pH

The formula for pH is \(pH =-\log[H^+]\). Substitute \([H^+]=0.00544\) into the formula:
\(pH=-\log(0.00544)\)
\(pH =-\log(5.44\times10^{-3})\)
Using the property of logarithms \(\log(ab)=\log a+\log b\), we get \(pH=-(\log5.44+\log10^{-3})\)
Since \(\log10^{-3}=- 3\) and \(\log5.44\approx0.735\)
\(pH=- (0.735 - 3)\)
\(pH=2.265\)

Answer:

\(2.27\)