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Answer:
0.0705 M
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O
From the balanced equation above,
The mole ratio of the acid, H₂SO₄ (nA) = 1
The mole ratio of the base, NaOH (nB) = 2
Next, data obtained from the question. This include the following:
Volume of acid, H₂SO₄ (Va) = 25 mL
Volume of base, NaOH (Vb) = 34.55 mL
Molarity of base, NaOH (Mb) = 0.1020 M
Molarity of acid, H₂SO₄ (Ma) =?
The molarity of the acid, H₂SO₄ can be obtained as follow:
MaVa / MbVb = nA/nB
Ma × 25 / 0.1020 × 34.55 = 1/2
Ma × 25 / 3.5241 = 1/2
Cross multiply
Ma × 25 × 2 = 3.5241
Ma × 50 = 3.5241
Divide both side by 50
Ma = 3.5241 / 50
Ma = 0.0705 M
Therefore, the concentration of the acid, H₂SO₄ is 0.0705 M
Answer:
C=0.140964M
Explanation:
M1V1=M2V2
DATA
M1=?
M2=0.1020M
V1=25.00ML
V2=34.55ML
M*25.00ML=(0.1020M)*(34.55ML)
M*25.00ML=3.5241M*ML
M=3.5241M*ML/25.00ML
M=0.140964M
Therefore, concentration of sulfuric acif is 0.140964M