Formula: % by mass = (mass of solute / mass of solution] *100
Data:
mass of solution = 80.85 g
% by mass = 22.4%
Unknown = mass of solute
Solution
% by mass = (mass of solute / mass of solution] *100 = >
mass of solute = % by mass * mass of solution / 100
mass of solute = 22.4 * 80.85 / 100 = 18.11 g
Answer: 18.11 g
You have to use the Henderson-Hasselbalch equation. Keep in mind that because the Pka is given the equation changes form slightly:
PH = Pka + log[acid/base]
Step 1 (Figure out the concentrations):
0.282 M of Acid (C6H5OOH) - 0.150 M = 0.132 M of acid
0.282 M of Base (C6HCOO) + 0.150 M = 0.432 M of bas3
Step 2 (Plug into equation):
PH = Pka + log[acid/base]
PH = 4.20 + log[0.132 M/0.432 M]
PH = 3.69
<u>Answer:</u> The hydroxide ion concentration and pOH of the solution is
and 2.88 respectively
<u>Explanation:</u>
We are given:
Concentration of barium hydroxide = 0.00066 M
The chemical equation for the dissociation of barium hydroxide follows:

1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions
pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution
To calculate pOH of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
![[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%282%5Ctimes%200.00066%29%3D1.32%5Ctimes%2010%5E%7B-3%7DM)
Putting values in above equation, we get:

Hence, the hydroxide ion concentration and pOH of the solution is
and 2.88 respectively
A. The concentration is in mol/L
Answer:
2) 0.4 mol
Explanation:
Step 1: Given data
- Volume of the solution (V): 500 mL
- Molar concentration of the solution (M): 0.8 M = 0.8 mol/L
Step 2: Convert "V" to L
We will use the conversion factor 1 L = 1000 mL.
500 mL × 1 L/1000 mL = 0.500 L
Step 3: Calculate the moles of KBr (solute)
The molarity is the quotient between the moles of solute (n) and the liters of solution.
M = n/V
n = M × V
n = 0.8 mol/L × 0.500 L = 0.4 mol