The new pH is 7.69.
According to Hendersen Hasselbach equation;
The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.
pH = pKa + log10 ([A–]/[HA])
Here, 100 mL of 0.10 m TRIS buffer pH 8.3
pka = 8.3
0.005 mol of TRIS.
∴ ![8.3 = 8.3 + log \frac{[0.005]}{[0.005]}](https://tex.z-dn.net/?f=8.3%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005%5D%7D%7B%5B0.005%5D%7D)
<em> </em>inverse log 0 = ![\frac{[B]}{[A]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
![\frac{[B]}{[A]} = 1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D%20%3D%201)
Given; 3.0 ml of 1.0 m hcl.
pka = 8.3
0.003 mol of HCL.
![pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69](https://tex.z-dn.net/?f=pH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005-0.003%5D%7D%7B%5B0.005%2B0.003%5D%7D%5C%5CpH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.002%5D%7D%7B%5B0.008%5D%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20log%20%7B0.25%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20%28-0.62%29%5C%5CpH%20%3D%207.69)
Therefore, the new pH is 7.69.
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Answer:
The final pressure of the gas is 0.915atm
Explanation:
We have to apply the Charles Gay Lussac Law, where the pressure changes directly proportional to absolute T°
- No change in volume
- The same moles in both situations
P1 / T1 = P2 / T2
0.991 atm / 342K = P2 / 316k
(0.991 atm / 342K) . 316K = P2
0.915 atm = P2
The amount of heat required to convert H₂O to steam is : 382.62 kJ
<u>Given data :</u>
Mass of liquid water ( m ) = 150 g
Temperature of liquid water = 43.5°C
Temperature of steam = 130°C
<h3 /><h3>Determine the amount of heat required </h3>
The amount of heat required = ∑ q1 + q2 + q3 ----- ( 1 )
where ;
q1 = heat required to change Temperature of water from 43.5°C to 100°C . q2 = heat required to change liquid water at 100°C to steam at 100°C
q3 = heat required to change temperature of steam at 100°C to 130°C
M* S
*ΔT
= 150 * 4.18 * ( 100 - 43.5 )
= 35425.5 J
moles * ΔHvap
= (150 / 18 )* 40.67 * 1000
= 338916.67 J
M * S
* ΔT
= 150 * 1.84 * ( 130 -100 )
= 8280 J
Back to equation ( 1 )
Amount of heat required = 35425.5 + 338916.67 + 8280 = 382622.17 J
≈ 382.62 kJ
Hence we can conclude that The amount of heat required to convert H₂O to steam is : 382.62 kJ.
Learn more about Specific heat of water : brainly.com/question/16559442
Answer:
its c.
Explanation:
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