Question

a 20.0 ml solution of naoh is neutralized with 51.1 ml of 0.200 m hbr.
what is the concentration of the original naoh solution?
answer:
m

Explanation:

Step1: Write the neutralization reaction

$$NaOH + HBr = NaBr + H_2O$$
From the reaction, the mole ratio of \(NaOH\) to \(HBr\) is \(1:1\).

Step2: Calculate the moles of \(HBr\)

Use the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, \(V\) is volume in liters).
\(V_{HBr}=51.1\space mL = 0.0511\space L\), \(C_{HBr} = 0.200\space M\)
\(n_{HBr}=C_{HBr}\times V_{HBr}=0.200\times0.0511 = 0.01022\space mol\)

Step3: Find the moles of \(NaOH\)

Since \(n_{NaOH}=n_{HBr}\) (from the \(1:1\) mole ratio), \(n_{NaOH}= 0.01022\space mol\)

Step4: Calculate the concentration of \(NaOH\)

\(V_{NaOH}=20.0\space mL=0.0200\space L\)
Use \(C=\frac{n}{V}\), \(C_{NaOH}=\frac{n_{NaOH}}{V_{NaOH}}=\frac{0.01022}{0.0200}=0.511\space M\)

Answer:

\(0.511\)