A buffer solution contains 0.475 M nitrous acid and 0.302 M sodium nitrite . If 0.0224 moles of potassium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution

Answers

Answer 1

Answer: The pH of the resulting solution will be 3.001

Explanation:

Molarity is calculated by using the equation:

[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex]          ......(1)

We are given:

Moles of NaOH = 0.0224 moles

Molarity of nitrous acid = 0.475 M

Molarity of sodium nitrite = 0.302 M

Volume of solution = 150 mL = 0.150 L          (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]\text{Moles of nitrous acid}=(0.475mol/L\times 0.150L)=0.07125mol[/tex]

[tex]\text{Moles of sodium nitrite}=(0.302mol/L\times 0.150L)=0.0453mol[/tex]

The chemical equation for the reaction of nitrous acid and NaOH follows:

                   [tex]HNO_2+NaOH\rightleftharpoons NaNO_2+H_2O[/tex]

I:                0.07125     0.0224     0.0453

C:             -0.0224     -0.0224   +0.0224

E:              0.04885         -          0.0677

The power of the acid dissociation constant is the negative logarithm of the acid dissociation constant. The equation used is:

[tex]pK_a=-\log K_a[/tex]        ......(2)

We know:

[tex]K_a[/tex] for nitrous acid = [tex]7.2\times 10^{-4}[/tex]

Using equation 2:

[tex]pK_a=-\log (7.2\times 10^{-4})=3.143[/tex]

To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

[tex]pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}[/tex]         .......(3)

Given values:

[tex][NaNO_2]=\frac{0.0677}{0.150}[/tex]

[tex][HNO_2]=\frac{0.04885}{0.150}[/tex]

[tex]pK_a=3.143[/tex]

Putting values in equation 3. we get:

[tex]pH=3.143-\log \frac{(0.0677/0.150)}{(0.04885/0.150)}\\\\pH=3.143-0.142\\\\pH=3.001[/tex]

Hence, the pH of the resulting solution will be 3.001


Related Questions

1. Determine the volume of SO2 (at STP) formed from the reaction of 96.7 mol FeS2 and 55.0 L of O2 at 358 K and 1.20 atm.
4 FeS2(s) + 11O2(g)  2Fe2O3(s) + 8SO2(g)

Answers

Answer:

40.0L of SO2 are produced

Explanation:

To solve this question we need to find the moles of O2 using PV = nRT in order to find the moles. Thus, we can find the limiting reactant and the moles (And volume) of SO2 produced as follows:

Moles O2:

n = PV/RT

n = 1.20atm*55.0L / 0.082atmL/molK*358K

n = 2.25 moles of O2.

Clearly, limiting reactant is O2.

The moles of SO2 produced are:

2.25 moles of O2 * (8mol SO2 / 11mol O2) = 1.6351 moles SO2

Volume SO2:

V = nRT/P

V = 1.6351 moles SO2*0.082atmL/molK*358K / 1.20atm

V = 40.0L of SO2 are produced

You are an intermediate product of an industrial process which intends to separate iron from its ore. A well known iron ore is hematite. Which of these ores does not contain iron?
Goethite
Malachite
Siderite
Limonite

Answers

Answer:

Malachite

Explanation:

Malachite is a copper carbonate hydroxide mineral, with the equation Cu2CO3(OH)2. This dark, green-joined mineral solidifies in the monoclinic precious stone framework, and frequently shapes botryoidal, sinewy, or stalagmitic masses, in cracks and profound, underground spaces, where the water table and aqueous liquids give the way to synthetic precipitation. So, the answer is malachite. Best of Luck!

Emily spills concentrated sodium hydroxide solution on her lab bench. What she should do first?

Answers

Answer:

Explanation: hell noo

Emily must notify the Instructor of the Lab/Classroom.

A diver exhales a bubble with volume of 250 mL at pressure of 2.4 atm and temperature of 15 C. How many gas particulate in this bubble?

Answers

Answer:

1.5x10²² particulates

Explanation:

Assuming ideal behaviour, we can solve this problem by using the PV=nRT formula, where:

P = 2.4 atmV = 250 mL ⇒ 250 / 1000 = 0.250 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 15 °C ⇒ 15 + 273 = 288 K

We input the given data:

2.4 atm * 0.250 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 288 K

And solve for n:

n = 0.025 mol

Finally we calculate how many particulates are there in 0.025 moles, using Avogadro's number:

0.025 mol * 6.023x10²³ particulates/mol = 1.5x10²² particulates

You are given a solution containing a pair of enantiomers (A and B). Careful measurements show that the solution contains 98% A and 2% B. What is the ee of this solution

Answers

Answer:

ee = 96%

Explanation:

Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the  differences of amounts of enantiomers and the total amunt. that is:

ee = |A-B|/ A+B * 100

ee = |98%-2%| / 98+2 * 100

ee = 96%

An infant acetaminophen suspension contains 80 mg/0.80 mL suspension. The recommended dose is 15 mg/kg body weight.
How many milliliters of this suspension should be given to an infant weighing 13 lb.

Answers

Answer:

0.8853 mL

Explanation:

First we convert 13 lb to kg, keeping in mind that 1 lb = 0.454 kg:

13 lb * [tex]\frac{0.454kg}{1lb}[/tex] = 5.902 kg

Then we calculate how many mg of acetaminophen should be given, using the recommended dose and infant mass:

15 mg/kg * 5.902 kg = 88.53 mg

Finally we calculate the required mL of suspension, using its concentration:

88.53 mg ÷ (80 mg/0.80 mL) = 0.8853 mL

Why does glucose and acentic acid have the same empirical formula

Answers

Answer:

Examples. Glucose (C6H12O6), ribose (C5H10O5), Acetic acid (C2H4O2), and formaldehyde (CH2O) all have different molecular formulas but the same empirical formula: CH2O.

Explanation:In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.

A 15.0 mL urine from a dehydrated patient has a density of 1.019g/mL. What is the mass of the sample, reported in mg?

Answers

Answer:

Mass of sample in mg = 15,285 mg

Explanation:

Given:

Volume of urine sample = 15 ml

Density of sample = 1.019 g/ml

FInd:

Mass of sample in mg

Computation:

Mass = density x volume

Mass of sample in mg = Volume of urine sample x Density of sample

Mass of sample in mg = 1.019 x 15

Mass of sample in mg = 15.285 gram

Mass of sample in mg = 15.285 x 1,000

Mass of sample in mg = 15,285 mg

Naturally occurring gallium is a mixture of isotopes
that contains 90.11% of Ga-69 (atomic mass = 68.93
u) and 9.89% of Ga-71 (atomic mass 70.92 u).
What is the average atomic mass of naturally
occurring gallium?
A) 69.93 amu
C) 69.50 amu
B) 69.12 amu
D) 69.00 amu

Answers

it is b have a great rest of your day

Can someone please help with these 2?

Answers

Equilibrium shifts to the right.

OPTION A

How much energy does an X-ray with an 8 nm (8 x 10-9m) wavelength have?
A. 1.99 x 10-25 J
B. 3.33 x 1016 J
C. 2.48 x 10-17 j
D. 8.28 x 10-26 J

Answers

Answer:

it would be option C

Explanation:

Speed of light = 3×10^8m/s

Planck's constant = 6.626×10^-34 Js

Wavelength = 8 x 10^-9 m

Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9

Energy = [19.878×10^(8-34)] / 8 x 10^-9

Energy = 2.48475 × 10^(-26+9)

Energy = 2.48×10^-17 J

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