A mass of 80 grams of Bromine would be
1 answer:
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Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O
Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O
Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L
∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion =![\frac{number of moles of [OH-]}{volume of solution (L)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bnumber%20of%20moles%20of%20%5BOH-%5D%7D%7Bvolume%20of%20solution%20%28L%29%7D%20)
= 
= 0.03199 M
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O
Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O
Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L
∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
= 0.0199 X 16.08 X 10^(-3)
= 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion =
Answer:
<em>The pH of the solution is 7.8</em>
Explanation:
The concentration of the solution is 0.001M and the dye could be in its protonated and deprotonated forms. If the concentration of the protonated form [HA] is 0.0002 M the concentration of the deprotonated form will be the subtraction between the concentration of the bye and the concentration of the protonated form:
[A-] = 0.001M - 0.0002M = 0.0008M
Also, the Henderson-Hasselbalch equation is
this equation shows the dependency between the pH of the solution, the pKa and the concentration of the protonated and deprotonated forms. Thus, replacing in the equation