Answer:
pKa = 3.51
Explanation:
The titration of acid solution with NaOH can be illustrated as:
Given that:
Volume of acid solution 
Volume of NaOH 
Molarity of acid solution 
Molarity of NaOH 
For Neutralization reaction:
Making 
the subject of the formula; we have:
However; since the number of moles of NaA formed is equal to the number of moles of NaOH used : Then :
Total Volume after titration = ( 25 + 18.8 ) m
= 43.8 mL
Molarity of salt (NaA ) solution = 
= 
= 0.0429 M
After mixing the two solution ; the volume of half neutralize solution is = 25 mL + 43.8 mL
= 68.8 mL
Molarity of NaA before mixing 
Volume 
Molarity of NaA after mixing 
Volume 
∴
Molarity of acid before mixing = 0.0725 M
Volume = 25 mL
Molarity of acid after mixing = 
= 0.0273 M
Since this is a buffer solution ; then using Henderson Hasselbalch Equation
![pH = pKa + log \frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
![3.51= pKa + log \frac{[0.0273]}{[0.0273]} \\ \\ 3.51= pKa + log \ 1 \\ \\ 3.51= pKa + 0 \\ \\ pKa = 3.51](https://tex.z-dn.net/?f=3.51%3D%20pKa%20%2B%20log%20%5Cfrac%7B%5B0.0273%5D%7D%7B%5B0.0273%5D%7D%20%5C%5C%20%20%5C%5C%203.51%3D%20pKa%20%2B%20log%20%5C%201%20%20%5C%5C%20%5C%5C%203.51%3D%20pKa%20%2B%200%20%5C%5C%20%5C%5C%20pKa%20%3D%203.51)
Answer:
<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>
Explanation:
This question is about solubility.
Regarding solubility, the solutions may be classified as:
- Unsaturated: the concentration is below the maximum concentration permited at the given temperature.
- Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.
- Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.
Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.
- In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.
- In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g of water.
- Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ
115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O ⇒ x = 57.5 g NaNO₃
- <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>
<em />
False reading may be obtained if there is a run over from adjacent reagent area in excessively wetted strips
Phenolphthalein only works efficiently from 4 to 10 ph
Litmus and phenolphthalein and methyl orange indicators can only allow you to tell if solution is alkaline,neutral or acidic
The two indicators with less limitations are methyl orange and thymol blue
Answer:
The strong forces oppose the electromagnetic force of repulsion between protons. Like ”glue” the strong force keeps the protons together to form the nucleus. The strong forces and electromagnetic forces both hold the atom together.
Explanation:
Hope This helps
