A negativley charged ion is the answer
Explanation:
Do the step 3 as outlined in the lab guide. record your results in the appropriate blank.
D
Answer:
all the below are ACIDS
Explanation:
1.HCOO- Formate
HCOO- is a conjugate base.
2.HNO3- Nitric Acid
Acid
3.CH3COOCH3: Methyl acetate
Acid
4. HCOOH: Formic Acid
Acid
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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.
Learn more about pH here:
brainly.com/question/24595796
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1. Has 4 sig figs, all non zeros are significant
2. 3 sig figs trailing zero after a decimal
3. 3 sig figs