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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Hey there!
Label A: Sublimation
Label B: Condensation
Label C: Melting
Remember sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state. Condensation is the conversion of a vapor or gas to a liquid. Melting is becoming liquefied by heat.
Hope this helps!
Bases
A base is a substance that dissociates into more hydroxide ions (-OH-) when dissolved in water. Bases are also good proton acceptors. Bases, therefore, reduce the number of H+ and increase OH- hence raising the pH of the solution.
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)
Explanation:
Other properties of bases is that they are bitter to the taste and they feel slippery when touched. Strong bases are nonthlese very corrosive like acids. Bases turn red litmus paper blue. Most alkali hydroxides such as NaOH are bases.
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Answer:
2HgS + 3O2 → 2HgO + 2SO2
The coefficients are: 2, 3, 2, 2
Explanation:
HgS + O2 → HgO + SO2
The equation can be balance as follow:
Put 3 in front of O2 as shown below:
HgS + 3O2 → HgO + SO2
Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:
HgS + 3O2 → 2HgO + 2SO2
Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:
2HgS + 3O2 → 2HgO + 2SO2
Now the equation is balanced.
The coefficients are: 2, 3, 2, 2
The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product
