81. Find the pH of each mixture of acids. a. 0.115 M in HBr and 0.125 M in HCHO2 b. 0.150 M in HNO2 and 0.085 M in HNO3 c. 0.185 M in HCHO2 and 0.225 M in HC2H3O2 d. 0.050 M in acetic acid and 0.050 M in hydrocyanic acid

See explanation below

Explanation:

This problem is a little long so I'm gonna be as clear as possible.

a) In this case we have two acids, HBr and HCHO2. Between these two acids, the HBr is the strongest, and does not have a Ka value to dissociate, while HCHO2 do.

In order to calculate pH we need the [H₃O⁺], and in this case, as HBr is stronger, the contribution of the weaker acid can be negligible, therefore, the pH of this mixture will be:

pH = -log[H₃O⁺]

pH = -log(0.115)

pH = 0.93

b) In this case it happens the same thing as part a) HNO₃ is the strongest acid, so the contribution of the HNO₂ which is a weak acid is negligible too, therefore the pH of this mixture will be:

pH = -log(0.085)

pH = 1.07

c) Now in this case, HCHO2 and HC2H3O2 are both weak acids, so to determine which is stronger, we need to see their Ka values. In the case of HCHO2 the Ka is 1.8x10⁻⁴ and for the HC2H3O2 the Ka is 1.8x10⁻⁵. Note that the difference between the two values of Ka is just 10¹ order, so, we can neglect the concentration of either the first or the second acid. We need to see the contribution of each acid, let's begin with the stronger acid first, which is the HCHO2, we will write an ICE chart to determine the value of the [H₃O⁺] and then, use this value to determine the same concentration for the second acid and finally the pH:

        HCHO₂ + H₂O <> CHO₂⁻ + H₃O⁺     Ka = 1.8*10⁻⁴

i)        0.185                                0          0

c)           -x                                 +x        +x

e)       0.185-x                             x           x

1.8*10⁻⁴ = x² / 0.185-x      

As Ka is small, we can assume that "x is small" too, therefore the (0.185-x) can be rounded to just 0.185 so:

1.8*10⁻⁴ = x²/0.185

1.8*10⁻⁴ * 0.185 = x²

x² = 3.33*10⁻⁵

x = 5.77*10⁻³ M = [H₃O⁺]

Now that we have this concentration, let's write an ICE chart for the other acid, but taking account this concentration of [H₃O⁺] as innitial in the chart, and solve for the new concentration of [H₃O⁺] (In this case i will use "y" instead of "x" to make a difference from the above):

        HC₂H₃O₂ + H₂O <> C₂H₃O₂⁻ + H₃O⁺       Ka = 1.8x10⁻⁵

i)          0.225                                  0           5.77x10⁻⁶

c)            -y                                     +y            +y

e)        0.225-y                                y           5-77x10⁻³+y

1.8x10⁻⁵ = y(5.77x10⁻³+y) / 0.225-y   ---> once again, y is small so:

1.8x10⁻⁵ = 5.77x10⁻³y + y² / 0.225

1.8x10⁻⁵ * 0.225 = 5.77x10⁻³y + y²

y² + 5.77x10⁻³y - 4.05x10⁻⁶ = 0

Solving for y:

y = -5.77x10⁻³ ±√(5.77x10⁻³)² + 4*4.05x10⁻⁶ / 2

y = -5.77x10⁻³ ±√4.95x10⁻⁵ / 2

y = -5.77x10⁻³ ± 7.04x10⁻³ / 2

y₁ = 6.35x10⁻⁴ M

y₂ = -6.41x10⁻³ M

We will take y₁ as the value, so the concentration of hydronium will be:

[H₃O⁺] = 5.77x10⁻³ + 6.35x10⁻⁴ = 6.41x10⁻³ M

Finally the pH for this mixture is:

pH = -log(6.41x10⁻³)

pH = 2.19

d) In this case, we have the same as part c, however the Ka values differ this time. The Ka for acetic acid is 1.8x10⁻⁵  while for HCN is 4.9x10⁻¹⁰. In this ocassion, we the difference in their ka is 10⁵ order, so we can neglect the HCN concentration and focus in the acetic acid. Let's do an ICE chart and then, with the hydronium concentration we will calculate pH:

         HC₂H₃O₂ + H₂O <> C₂H₃O₂⁻ + H₃O⁺       Ka = 1.8x10⁻⁵

i)          0.050                                  0              0

c)            -y                                     +y            +y

e)        0.050-y                                y              y

1.8*10⁻⁵ = y² / 0.050-y      

As Ka is small, we can assume that "y is small" too

1.8*10⁻⁵ = y²/0.050

1.8*10⁻⁵ * 0.050 = y²

y² = 9*10⁻⁷

y = 9.45*10⁻⁵ M = [H₃O⁺]

Finally the pH:

pH = -log(9.45x10⁻⁵)

pH = 3.02

the dissolution of polymer can be thermodynamically described in 2 steps.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.