Perchloric acid (HCO₄) is one of strongest inorganic acids. Perchloric acid precipitation is used to removes most of the protein present in the
sample and stabilize many of the small molecule analytes. It can use also to precipitation glycogen, ATP, glutathione, antioxidants.
Perchloric acid donates protons to macromolecules and lower their solubility in water. Also perchloric acid is strong oxidizing agent and change oxidation number of elements.
Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:
![H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O](https://tex.z-dn.net/?f=H_3PO_4%20%5C%20%2B%20%5C%203NaOH%20-----%3E%20Na_3%20PO_4%20%5C%20%2B%20%5C%203H_2O)
1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows
![H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O](https://tex.z-dn.net/?f=H_3PO_4%20%5C%20%2B%20%5C%203NaOH%20-----%3E%20Na_3%20PO_4%20%5C%20%2B%20%5C%203H_2O)
10 ml 17.50 ml
(x) M 0.200 M
Molarity = ![\frac{0.2*17.5}{1000}](https://tex.z-dn.net/?f=%5Cfrac%7B0.2%2A17.5%7D%7B1000%7D)
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
= ![0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}](https://tex.z-dn.net/?f=0.0035%20%5C%20mole%20%5C%20of%20NaOH%2A%20%5Cfrac%7B1%20mole%20of%20H_3PO_4%7D%7B3%20%5C%20mole%20%5C%20of%20%5C%20NaOH%7D)
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration = ![\frac{mole \ \ of \ soulte }{ Volume \ of \ solution }](https://tex.z-dn.net/?f=%5Cfrac%7Bmole%20%5C%20%5C%20of%20%5C%20soulte%20%7D%7B%20Volume%20%5C%20of%20%5C%20solution%20%7D)
Molar Concentration = ![\frac{0.00166 \ mole \ of \ H_3PO_4 }{10}*1000](https://tex.z-dn.net/?f=%5Cfrac%7B0.00166%20%5C%20mole%20%5C%20of%20%5C%20%20H_3PO_4%20%7D%7B10%7D%2A1000)
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M
Answer:
Methods for determining or delivering precise volumes include volumetric pipets and pycnometers; less precise methods include burets, graduated cylinders, and graduated pipets. In this experiment, you will measure masses and volumes to determine density. Four different metal cylinders are investigated.
Explanation:
Answer:
Explanation:
By definition, <em>half neutralization</em> is the point at which half of the acid has been neutralized.
The neutralization reaction that you are studying is the acid-base reaction:
- HCl (aq) + NaOH (aq) → NaCl(aq) + H₂O (aq)
Then, since the starting molarity of the acid (HCl) is 0.2 M, you just need to find half of that concentration:
- Half molarity = M / 2 = 0.2 M / 2 = 0.1 M
So, the answer is the first choice: a. 0.1 M.