Question

1. A 100 mL solution of 0.200 M HF is titrated with 0.100 M Ba(OH)₂. What is the volume of Ba(OH)₂ needed to reach equivalence point?

200 mL

50 mL

100 mL

300 mL

Cannot determine based on the provided information.

2. A 100 mL solution of 0.200 M NH₃ is titrated with 0.100 M HCl. What is the volume of HCl needed to reach equivalence point?

200 mL

100 mL

50 mL

300 mL

Cannot determine based on the provided information.

3. A 100 mL solution of 0.200 M NH₃ is titrated with 0.100 M HF. What is the volume of HF needed to reach equivalence point?

100 mL

50 mL

300 mL

Cannot determine based on the provided information.

4.      A 100 mL solution of 0.200 M Sr(OH)₂ is titrated with 0.100 M H₃PO₄. What is the volume of H₃PO₄ needed to reach equivalence point?

100 mL

200 mL

50 mL

300 mL

133 mL

5. A 100 mL solution of unknown concentration H₂SO₄ is titrated with 200 mL of 0.100 M Ba(OH)₂ solution to reach equivalence point. What is the concentration of H₂SO₄ solution?

0.100 M

0.050 M

0.020 M

0.200 M

Cannot determine based on the provided information.

6. A 100 mL solution of unknown concentration H₃PO₄ is titrated with 200 mL of 0.100 M Sr(OH)₂ solution to reach equivalence point. What is the concentration of H₃PO₄ solution?

0.100 M

0.133 M

0.200 M

0.050 M

0.300 M

A 100 mL solution of 0.200 M HF(K_a=6.6*10^-4) is titrated with 0.100 M Ba(OH)₂. It takes 100 mL of Ba(OH)₂ to reach the equivalence point. What is the pH at half-equivalence point?

7

3.5

3.18

10.5

10.82

8. Based on your answer in #7, what is the concentration of H₃O⁺ at half-equivalence point?

6.6 × 10^-4 M

6.61 × 10^-4 M

10^-7 M

3.2 × 10^-4 M

3.16 × 10^-4 M

9. A 100 mL solution of 0.200 M unknown weak acid is titrated with 0.100 M NaOH. The pH at half-equivalence point is 5.52. What is the K_a value for the unknown weak acid?

3.02 × 10^-6

3.31 × 10⁵

3.0 × 10^-6

3.3 × 10⁵

Cannot determine based on the provided information.

10. Based on your answer in #9, what is the concentration of H₃O⁺ at half-equivalence point?

3.0 × 10^-6

3.02 × 10^-6

3.31 × 10⁵

3.3 × 10⁵

Cannot determine based on the provided information

Answer 1

1)

Balanced chemical equation is:

2 HF + Ba(OH)2 ---> BaF2 + 2 H2O

Here:

M(HF)=0.2 M

M(Ba(OH)2)=0.1 M

V(HF)=100.0 mL

According to balanced reaction:

1*number of mol of HF =2*number of mol of Ba(OH)2

1*M(HF)*V(HF) =2*M(Ba(OH)2)*V(Ba(OH)2)

1*0.2 M *100.0 mL = 2*0.1M *V(Ba(OH)2)

V(Ba(OH)2) = 100 mL

Answer: 100 mL

2)

Balanced chemical equation is:

NH3 + HCl ---> NH4Cl +

Here:

M(NH3)=0.2 M

M(HCl)=0.1 M

V(NH3)=100.0 mL

According to balanced reaction:

1*number of mol of NH3 =1*number of mol of HCl

1*M(NH3)*V(NH3) =1*M(HCl)*V(HCl)

1*0.2 M *100.0 mL = 1*0.1M *V(HCl)

V(HCl) = 200 mL

Answer: 200 mL

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