What volume (in mL) of a 0.341 M HCl solution is needed to neutralize 43.9 mL of a 0.480 M NaOH solution

61.8 mL

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

HCl + NaOH —> NaCl + H2O

From the balanced equation above,

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Next, the data obtained from the question include:

Molarity of acid, HCl (M1) = 0.341 M

Volume of base, NaOH (Vb) = 43.9 mL

Molarity of base, NaOH (Mb) = 0.480 M

Volume of acid, HCl (Va) =?

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

MaVa/MbVb = nA/nB

0.341 × Va / 0.480 × 43.9 = 1/1

Cross multiply

0.341 × Va × 1 = 0.480 × 43.9 × 1

0.341 × Va = 21.072

Divide both side by 0.341

Va = 21.072 /0.341

Va = 61.8 mL

Therefore, 61.8 mL of the acid, HCl is needed for the reaction.

it's all four options.

can you list the answers so i can answer