Question

question 16 (5 points)
what is the concentration of an hbr solution if 12.0 ml of the solution is neutralized by 15.0 ml of a 0.25 m koh solution?

○ a) 0.75 m

○ b) 0.31 m

○ c) 0.40 m

○ d) 0.20 m

Explanation:

Step1: Recall neutralization formula

For acid - base neutralization, \(n_{acid}=n_{base}\). The formula for moles is \(n = M\times V\) (where \(M\) is molarity and \(V\) is volume in liters). For \(HBr\) (acid) and \(KOH\) (base), the reaction is \(HBr + KOH=KBr + H_2O\), so the mole ratio of \(HBr\) to \(KOH\) is \(1:1\). Thus, \(M_{HBr}\times V_{HBr}=M_{KOH}\times V_{KOH}\).

Step2: Convert volumes to liters (or use mL directly as units cancel)

\(V_{HBr} = 12.0\space mL\), \(V_{KOH}=15.0\space mL\), \(M_{KOH} = 0.25\space M\). Rearranging the formula for \(M_{HBr}\): \(M_{HBr}=\frac{M_{KOH}\times V_{KOH}}{V_{HBr}}\)

Step3: Substitute values

\(M_{HBr}=\frac{0.25\space M\times15.0\space mL}{12.0\space mL}\)
\(M_{HBr}=\frac{3.75}{12.0}\space M = 0.3125\space M\approx0.31\space M\)

Answer:

b) 0.31 M