Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54
Answer:
the properties of catalyst are
it remains unchanged after chemical reactions
it accelerate or deaccelerate the reaction without taking part in it
Explanation:
they are used to convert raw materials into useful one
catalyst are integral in making plastics
<u>36 ml of NaOh and</u><u> 464 ml</u><u> of </u><u>HCOOH</u><u> would be enough to form 500 ml of a buffer with the same pH as the buffer made with </u><u>benzoic acid </u><u>and NaOH.</u>
What is benzoic acid found in?
- Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
- Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.
Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.
Amount of moles of NaOH -2 × 0.025 = 0.05 mol
Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol
In this case, we can calculate the pH produced by the buffer of these two reagents, as follows
We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows
Now we must solve the equation above. This will be done using the following values
With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.
NaOH volume
( 0.5 - 0.464)L
0.036L .................... 36ml
HCOOH volume
500 - 36 = 464mL
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The molecular mass of the immunoglobulin G, given the data from the question is 1.53×10⁵ g/mole
<h3>How to determine the molarity</h3>
We'll begin by calculating the molarity of the immunoglobulin G. This is illustrated below:
- Volume = 0.106 L
- Temperature (T) = 25 °C = 25 + 273 = 298 K
- Osmotic pressure (π) = 0.733 mbar = 0.733 × 0.000987 = 0.00072 atm
- Gas constant (R) = 0.0821 atm.L/Kmol
- Van't Hoff factor (i) = 1
- Molarity (M)
π = iMRT
M = π / iRT
M = 0.00072 / (1 × 0.0821 × 298)
M = 0.000029 M
<h3>How to determine the mole of immunoglobulin G</h3>
- Molarity = 0.000029 M
- Volume = 0.106 L
- Mole =?
Mole = Molarity × volume
Mole = 0.000029 × 0.106
Mole = 3.074×10⁻⁶ mole
<h3>How to determine the molar mass of mmunoglobulin G</h3>
- Mole = 3.074×10⁻⁶ mole
- Mass = 0.470 g
- Molar mass =?
Molar mass = mass / mole
Molar mass = 0.47 / 3.074×10⁻⁶
Molar mass = 1.53×10⁵ g/mole
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The answer is for this question is c
