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)?
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Ответ:
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.
Ответ: