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What volume (in mL) of 0.2600 M HBr is

required to neutralize 40.00 mL of 0.8000 M

NaOH?

Sagot :

Eduard22sly

The volume of 0.26 M HBr required to neutralize 40 mL of 0.8 M NaOH is 123.08 mL

We'll begin by writing 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, HBr (nA) = 1

The mole ratio of the base, NaOH (nB) = 1

From the question given above, the following data were:

Volume of base, NaOH (Vb) = 40 mL

Molarity of base, NaOH (Mb) = 0.8 M

Molarity of acid, HBr (Ma) = 0.26 M

Volume of acid, HBr (Va) =?

The volume of the acid, HBr needed for the reaction can be obtained as follow:

MaVa / MbVb = nA / nB

0.26 × Va / (0.8 × 40) = 1

0.26 × Va / 32 = 1

Cross multiply

0.26 × Va = 32

Divide both side by 0.26

Va = 32 / 0.26

Va = 123.08 mL

Therefore, the volume the acid, HBr needed to neutralize the base, NaOH is 123.08 mL

Learn more: https://brainly.com/question/9659507

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