To get the molarity, you divide the moles of solute by the litres of solution.
Molarity
=
moles of solute
litres of solution
For example, a 0.25 mol/L NaOH solution contains 0.25 mol of sodium hydroxide in every litre of solution.
To calculate the molarity of a solution, you need to know the number of moles of solute and the total volume of the solution.
To calculate molarity:
Calculate the number of moles of solute present.
Calculate the number of litres of solution present.
Divide the number of moles of solute by the number of litres of solution.
Answer:
Use either a vinegar, Borax, or bleach solution in a spray bottle to tackle the mold. Simply spray the solution on showers, baths, basins, tiles, grout, or caulking. Then use either a cleaning cloth or a toothbrush to remove the slime
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
![[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%20%5D%20%3D%20%5Csqrt%7BCa%20%5Ctimes%20Ka%20%7D%20%3D%20%5Csqrt%7B0.249%20%5Ctimes%204.50%20%5Ctimes%2010%5E%7B-4%7D%20%20%7D%20%3D%200.0106%20M)
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
![\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D%20%5Ctimes%20100%5C%25%20%3D%20%5Cfrac%7B0.0106M%7D%7B0.249%7D%20%5Ctimes%20100%5C%25%20%3D%204.26%5C%25)
Buffers - mixtures of conjugate acid and conjugate base at ±1 pH unit from pH = pKa. Resistant to changes in pH in response to small additions of H+ or OH-. ... Polyprotic acids - dissociation of each H+ can be treated separately if the pKa values are different
