Chem SL/HL MC Help

[Guest]

1) How does the pH of the mixture change as NaOH is slowly added to a solution of HCl?

• The pH increases and may rise above 7
• The pH decreases and may fall below 7
• The pH increases to 7 and stops
• The oh decreases to 7 and stops
• The pH remains at 7

Answer: The pH increases and may rise above 7. How?

 

2) To 100mL of 1.0 M aqueous HCl, enough water is added to make 1.0 L of solution. During this dilution, the pH

• falls by about 1 pH unit
• falls by about 10 unit
• changes by about 9 units
• rises by about 1 unit
• rises by abouit 10 units

Answer: rises by about 1 pH unit. How, why?

 

3) What is the concentration of hydroxide ions in a solution of 0.350 mol/L HI?

• 0.350 mol/L
• 0.175 mol/L
• 2.86 x 10^-14
• 0.650 mol/L
• 1.30 mol/L

Answer: 2.86 x 10^-14   How?

 

4) A solution is made up of 25 mL of 0.250 mol/L H2SO4 and 65 mL of 0.050 mol/L NaOH. What is the concentration of hyrdonium ions in the solution?

• 0.25 mol/L
• 0.051 mol/L
• 0.10 mol/L
• 0.033 mol/L
• 0.066 mol/L

Answer: 0.051 mol/L

Edited November 12, 2015 by frank!e

[kw0573]

1) Adding a base would increase the pH. Eventually the pH would rise above 7 (neutral solution).

If you are not yet comfortable with this concept, you should definitely re-read the textbook and review your notes before making further progress in the acids/base unit. I don't think you are ready for the other three questions yet.

[Msj Chem]

1) By adding a base, you are increasing the concentration of OH^- (hydroxide ions). These react with H⁺ (hydronium ions) to form H₂O (water). 

Therefore the pH will increase (as concentration of H⁺ decreases). If the concentration of OH^- ions is greater than the concentration of H⁺ ions, the pH will be greater than 7. 

 

2) By diluting the acid by a factor of 10 (from 100 cm³ to 1000 cm³), you are reducing the [H⁺] also by a factor of 10. 

The pH scale is a logarithmic scale which means that a change in one unit of pH represents a 10 x change in the [H⁺].

Because the [H⁺] is decreased by a factor of 10, then the pH increases by one unit. 

 

3) HI is a strong acid so dissociates completely in solution.

A 0.350 moldm^-3 solution will produce 0.350 moldm^-3 [H⁺]

At 298 K, [H⁺] x [OH^-] = 1.00 x 10^-14

[OH^-] = 1.00 x 10^-14 / 0.350 = 2.86 x 10^-14

 

4) H₂SO₄ + 2NaOH ==> Na₂SO₄ + 2H₂O

n= CV 

H₂SO₄ 0.250 x (25/1000) = 6.25 x 10^-3 mol ( x 2 because H₂SO₄ is diprotic) = 0.0125 mol H⁺ 

NaOH 0.050 x (65/1000) = 3.25 x 10^-3 mol OH^-

0.0125 - 3.25 x 10^-3 = 9.25 x 10^-3 

[H⁺] = 9.25 x 10^-3  / (90/1000) = 0.10 mol dm^-3

Edited November 2, 2015 by Msj Chem

[Guest]

1) By adding a base, you are increasing the concentration of OH^- (hydroxide ions). These react with H⁺ (hydronium ions) to form H₂O (water). 

Therefore the pH will increase (as concentration of H⁺ decreases). If the concentration of OH^- ions is greater than the concentration of H⁺ ions, the pH will be greater than 7. 

 

2) By diluting the acid by a factor of 10 (from 100 cm³ to 1000 cm³), you are reducing the [H⁺] also by a factor of 10. 

The pH scale is a logarithmic scale which means that a change in one unit of pH represents a 10 x change in the [H⁺].

Because the [H⁺] is decreased by a factor of 10, then the pH increases by one unit. 

 

3) HI is a strong acid so dissociates completely in solution.

A 0.350 moldm^-3 solution will produce 0.350 moldm^-3 [H⁺]

At 298 K, [H⁺] x [OH^-] = 1.00 x 10^-14

[OH^-] = 1.00 x 10^-14 / 0.350 = 2.86 x 10^-14

 

4) H₂SO₄ + 2NaOH ==> Na₂SO₄ + 2H₂O

n= CV 

H₂SO₄ 0.250 x (25/1000) = 6.25 x 10^-3 mol ( x 2 because H₂SO₄ is diprotic) = 0.0125 mol H⁺ 

NaOH 0.050 x (65/1000) = 3.25 x 10^-3 mol OH^-

0.0125 - 3.25 x 10^-3 = 9.25 x 10^-3 

[H⁺] = 9.25 x 10^-3  / (90/1000) = 0.10 mol dm^-3

For #4, the answer is 0.051 M, not 0.010M. So would I just do the same thing? Also, where did you get 90/1000?

Edited November 12, 2015 by frank!e