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