Question

calculate the ph of a 0.00084 m solution of hbr at 25.0 ^{circ}c.

Explanation:

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

Since \(HBr\) is a strong acid, it completely dissociates in water: \(HBr
ightarrow H^++Br^-\). So, \([H^+]=[HBr] = 0.000884\space M\)

Step2: Calculate the \(pH\)

The formula for \(pH\) is \(pH=-\log[H^+]\). Substitute \([H^+]=8.84\times 10^{-4}\) into the formula:
\(pH =-\log(8.84\times 10^{- 4})\)
Using the logarithm property \(\log(ab)=\log a+\log b\), we have \(pH=-(\log8.84+\log(10^{-4}))\)
\(\log8.84\approx0.946\) and \(\log(10^{-4})=- 4\)
\(pH=- (0.946-4)=3.054\)

Answer:

\(3.05\)