The first law of thermodynamics states that ΔE=Q-W. How is this also a statement of the principle of conservation of energy?

Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result

Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f) 12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) = 4.886

(d) mol HA reacted = 150 x 10⁻³ L x x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) = 5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH = √(kb x (A⁻) where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA, 200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒ Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0

3 years ago

What is the oh- in a solution with a poh of 5.71

Rudik [331]

Answer:- The hydroxide ion concentration of the solution is $1.95*10^-^6$ .

Solution:- The formula used to calculate pOH from hydroxide ion is:

$pOH=-log[OH^-]$

When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:

$[OH^-]=10^-^p^O^H$

pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:

$[OH^-]=10^-^5^.^7^1$

$[OH^-]$ = 0.00000195 or $1.95*10^-^6$

So, the hydroxide ion concentration of the solution is $1.95*10^-^6$ .

3 0

3 years ago

How do you calculate mass using density and volume?

skad [1K]

Example:

Mass = ?

Density = 25 g/mL

Volume = 5 mL

therefore:

d = m / V

25 = m / 5

m = 25 x 5

m = 125 g

hope this helps!

7 0

3 years ago

A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a

svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = $62.364\text{ L. Torr }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol$

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g$

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0

3 years ago

EXTRA CREDIT. The rate at which an object moves is its speed. If a horse

Irina18 [472]

Average speed is total distance divided by total time.

Total distance is 50 m + 150 m + 300 m = 500 m
Total time is 68 s + 35 s + 22 s = 125 s

Average speed is 500 m / 125 s = 4 m/s

6 0

3 years ago