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
![pH = pKa + Log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20Log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
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
Answer:
1.03
Explanation:
You would take 0.00103 and move the decimal like this; 0001.03, we wouldn't have the zeroes in front of the one, as it can throw us off. Therefore, your answer would be 1.03.
Hope I helped! Don't hesitate to let me know if I made a mistake.
Answer:
0.0303 Liters
Explanation:
Given:
Mass of the potassium hydrogen phosphate = 0.2352
Molarity of the HNO₃ Solution = 0.08892 M
Now,
From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃
The number of moles of 0.2352 g of potassium hydrogen phosphate
= Mass / Molar mass
also,
Molar mass of potassium hydrogen phosphate
= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol
Number of moles = 0.2352 / 174.15 = 0.00135 moles
thus,
The number of moles of HNO₃ required for 0.00135 moles
= 2 × 0.00135 mol of HNO₃
= 0.0027 mol of HNO₃
Now,
Molarity = Number of Moles / Volume
thus,
for 0.0027 mol of HNO₃, we have
0.08892 = 0.0027 / Volume
or
Volume = 0.0303 Liters
