A chemistry student named Igor was performing a titration experiment. He used 500mL of 6M sulfuric acid, H2SO4(aq) to completely neutralize a 2.00 L solution of unknown concentration sodium hydrodixe, NaOH(aq). What is the concentration of the NaOH(aq) in molarity (M)?

3 M

Explanation:

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

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

From the balanced equation above, the following data were obtained:

Mole ratio of the acid, H2SO4 (nA) = 1

Mole ratio of the base, NaOH (nB) = 2

Next, the data obtained from the question. This include the following:

Volume of acid, H2SO4 (Va) = 500 mL

Molarity of acid, H2SO4 (Ma) = 6 M

Volume of base, NaOH (Vb) = 2 L

Molarity of base, NaOH (Mb) =?

Next, we shall convert 2 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

2 L = 2 L × 1000 mL /1 L

2 L = 2000 mL

Thus, 2 L is equivalent to 2000 mL.

Finally, we shall determine the molarity of the base, NaOH. This can be obtained as follow:

Volume of acid, H2SO4 (Va) = 500 mL

Molarity of acid, H2SO4 (Ma) = 6 M

Volume of base, NaOH (Vb) = 2000 mL

Mole ratio of the acid, H2SO4 (nA) = 1

Mole ratio of the base, NaOH (nB) = 2

Molarity of base, NaOH (Mb) =?

MaVa / M Vb = nA/nB

6 × 500 / Mb × 2000 = 1/2

3000 / Mb × 2000 = 1/2

Cross multiply

Mb × 2000 × 1 = 3000 × 2

Mb × 2000 = 6000

Divide both side by 2000

Mb = 6000 /2000

Mb = 3 M

Thus, the molarity of the base, NaOH is 3 M.

right

explanation: