Question

3. A volume of 90 mL of 0.2 M HBr neutralizes
a 60 mL sample of NaOH solution. What is
the concentration of the NaOH solution?

Answer 1

Answer:

0.3M

Explanation:

Step 1:

Data obtained from the question. This include the followingb:

Volume of acid (Va) = 90mL

Concentration of acid (Ca) = 0.2M

Volume of base (Vb) = 60mL

Concentration of base (Cb) =....?

Step 2:

The balanced equation for the reaction. This is given below:

HBr + NaOH —> NaBr + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 3:

Determination of the concentration of the base, NaOH.

The concentration of the base can be obtained as follow:

CaVa /CbVb = nA/nB

0.2 x 90 / Cb x 60 = 1

Cross multiply

Cb x 60 = 0.2 x 90

Divide both side by 60

Cb = 0.2 x 90 /60

Cb = 0.3M

Therefore, the concentration of the base, NaOH is 0.3M

Answer 2

Answer:

$M_{NaOH}=0.3M$

Explanation:

Hello,

In this case, since we are talking about a neutralization reaction, the moles of acid must equal the moles of base as shown below:

$n_{HBr}=n_{NaOH}$

Thus, in terms of molarities we've got:

$M_{HBr}V_{HBr}=M_{NaOH}V_{NaOH}$

This is possible since HBr reacts with NaOH in a 1:1 molar ratio:

$HBr+NaOH\rightarrow NaBr+H_2O$

Hence, for the given concentration and volume of hydrobromic acid and the volume of sodium hydroxide, we compute its concentration as shown below:

$M_{NaOH}=\frac{M_{HBr}V_{HBr}}{V_{NaOH}} =\frac{90mL*0.2M}{60mL} \\\\M_{NaOH}=0.3M$

Best regards.