Question

CHEM HELP PLS 30 POINTS!!

1. what volume of 0.20 M HCl (aq) is needed to titrate 50 mL of 0.10 M NaOH to the endpoint?
A) 10.0 mL HCl
B) 40.0 mL HCl
C) 25.0 mL HCl
D) 20.0 mL HCl

2. what is the molarity of a NaOH solution if 40 mL of the solution is titrated to the endpoint with 15 mL of 1.50 M H2SO4?
A) 1.13 M
B) 1.47 M
C) 0.56 M
D) 0.75 M

3. suppose that 10.0 mL of HNO3 is neutralized by 71.4 mL of a 4.2 x 10 ^ -3 M solution of KOH in a titration. Calculate the concentration of the HNO3 solution.
A) 1.0 x 10 ^ -2 M
B) 3.0 x 10 ^ -2 M
C) 5.5 x 10 ^ -2 M
D) 1.0 x 10 ^ -14 M

Answer 1

1. 25 ml
2. 0.56 M
3. 3.0x 10^-2 M
Step by step explanation as follow
Use the formula (Molarity x volume) = ( molarity x volume)

Answer 2

1. The volume of HCl needed is 25 mL
2. The molarity of NaOH needed is 1.13 M
3. The concentration of HNO₃ is 3×10¯² M

1. How to determine the volume of HCl needed

Balanced equation

HCl + NaOH —> NaCl + H₂O

From the balanced equation above,

• The mole ratio of the acid, HCl (nA) = 1
• The mole ratio of the base, NaOH (nB) = 1

From the question given above, the following data were obtained:

• Volume of base, NaOH (Vb) = 50 mL
• Concentration of base, NaOH (Cb) = 0.1 M
• Concentration of acid, HCl (Ca) = 0.2
• Volume of acid, HCl (Va) =?

CaVa / CbVb = nA / nB

(0.2 × Va) / (0.1 × 50) = 1

(0.2 × Va / 5  = 1

Cross multiply

0.2 × Va  = 5

Divide both side by 0.2

Va = 5 / 0.2

Va = 25 mL (Option C)

2. How to determine the molarity of NaOH

Balanced equation

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

From the question given above, the following data were obtained:

• Volume of base, NaOH (Vb) = 40 mL
• Concentration of acid, H₂SO₄ (Ca) = 1.5 M
• Volume of acid, H₂SO₄ (Va) = 15 mL
• Concentration of base, NaOH (Cb) =?

CaVa / CbVb = nA / nB

(1.5 × 15) / (Cb × 40) = 1 / 2

22.5 / (Cb × 40) = 1 / 2

Cross multiply

Cb × 40 =  22.5 × 2

Divide both side by 40

Cb = (22.5 × 2) / 40

Cb = 1.13 M (Option A)

3. How to determine the molarity of HNO₃

Balanced equation

HNO₃ + KOH —> KNO₃ + H₂O

From the balanced equation above,

• The mole ratio of the acid, HNO₃ (nA) = 1
• The mole ratio of the base, KOH (nB) = 1

From the question given above, the following data were obtained:

• Volume of base, KOH (Vb) = 71.4 mL
• Concentration of base, KOH (Cb) = 4.2×10¯³ M
• Volume of acid, HNO₃ (Va) = 10 mL
• Concentration of acid, HNO₃ (Ca) =?

CaVa / CbVb = nA / nB

(Ca × 10) / (4.2×10¯³ × 71.4) = 1

(Ca × 10) / 0.29988 = 1

Cross multiply

Ca × 10 = 0.29988

Divide both side by 10

Ca = 0.29988 / 10

Ca = 3×10¯² M (Option B)

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