All categories

3 years ago

Use the following information to calculate the concentration, Ka and pka for an unknown monoprotic weak acid. (8 pts.) 20.00 mL

Chemistry

1 answer:

lisov135 [29]3 years ago

6 0

Answer:

Concentration: 0.185M HX

Ka = 9.836x10⁻⁶

pKa = 5.01

Explanation:

A weak acid, HX, reacts with NaOH as follows:

HX + NaOH → NaX + H2O

<em>Where 1 mole of HX reacts with 1 mole of NaOH</em>

To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).

18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX

In 20.0mL = 0.0200L =

0.00370 moles HX / 0.0200L = 0.185M HX

The equilibrium of HX is:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

And Ka is defined as:

Ka = [H⁺] [X⁻] / [HX]

<em>Where [H⁺] = [X⁻] because comes from the same equilibrium</em>

As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M

Replacing:

Ka = [H⁺] [H⁺] / [HX]

Ka = [1.349x10⁻³M]² / [0.185M]

Ka = 9.836x10⁻⁶

pKa = -log Ka

You might be interested in

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0

andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = $-\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B$

If $C_{A0}$ and $C_{B0}$ are the inital concentrations and x the concentration reacted at time t, so $C_A=C_{A0} -x$ and $C_B=C_{B0} -x$ and the rate at time t is written as:

$-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)$

Separating variables and integrating

$\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt$

The integral in left side is solved by partial fractions, it can be used integral tables

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt$

Using logarithm properties (ln x - ln y = ln(x/y))

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a $C_{B0}$ of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, $C_{A0}=0.075$ , $C_{B0}=0.05$ , $C_{B}=0.02$ , it implies that the quantity reacted, x, is 0.03 and $C_{A}=0.075-x = 0.045$ . Then, the value of k would be

$kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})$

$k = 16.21 \frac{dm^3}{mol*1h}$

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

$\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt$

but suppossing that $C_A= C_{A0}/2=0.0375$ so $C_B=C_{B0}- C_{A0}/2=0.0125$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})$

$t=1.71 h = 1.71*3600 s = 6.1*10^3 s$

For reactant B, $C_B= C_{B0}/2=0.025$ so $C_A=C_{A0}- C_{B0}/2=0.05$ , k=16.2 and the same initial concentrations. Replacing in the equation

$t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})$

$t=0.71 h = 0.7*3600 s = 2.5*10^3 s$

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0

2 years ago

How does Kinetic energy and how does it relates to diffusion and temperature.

olya-2409 [2.1K]

Answer:

Explanation:

khfdnkv Klein pick pick or en la Web outta

5 0

3 years ago

How many calories of heat are needed to raise the temperature of 33.0 g of diethyl ether from 14.0 ∘C to 25.0 ∘C ?

Xelga [282]

The amount of heat (in calories) needed to raise the temperature is 324.885 calories

<h3>How to determine the temperature change </h3>

We'll begin by obtaining the temperature change. This can be obtained as followed:

Initial temperature (T₁) = 14 °C
Final temperature (T₂) = 25 °C
Change in temperature (ΔT) = ?

ΔT = T₂ - T₁

ΔT = 25 - 14

ΔT = 11 °C

<h3>How to determine the heat (in Calories)</h3>

The amount of heat needed to raise the temperature can bee obtaimedals follow:

Mass (M) = 33 g
Change in temperature (ΔT) = 11 °C
Specific heat capacity (C) of diethyl ether = 0.895 cal/g°C
Heat (Q) =?

Q = MCΔT

Q = 33 × 0.895 × 11

Q = 324.885 calories

Thus, the amount of heat needed is 324.885 calories

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

3 0

1 year ago

Write any four system of measurement

Elena L [17]

Answer:

Systems of measurement in use include the International System of Units (SI), the modern form of the metric system, the imperial system, and United States customary units.

5 0

3 years ago

Read 2 more answers

PLSS HELPPP ILL GIVE 30 POINTS PLSS its for a test plss

alina1380 [7]

Answer:

I am sorry but this is cheating you have to come up with your own answer

Explanation:

5 0

3 years ago