To clean a clogged drain, 26 g of sodium hydroxide is added to water to make 150 mL of solution. What are the pH and pOH values for the solution?
NaOH is a strong base. (NaOH) = (OH^-).
pOH = -log(OH^-)
Post your work if you get stuck.
You can't do that^ because you get a pH of over 14. Anyone else know how to get the answer? I heard somewhere that it was 11.6, how do you get that?
You can still get a pH over 14, those are not the limits. you can also have negative numbers. i can't really explain how or why, i just know that its possible. but i was replying to Joe, because 14 is not the cut off point.
To determine the pH and pOH values of a solution, you need to know the concentration of hydroxide ions (OH-) or hydrogen ions (H+). In this case, we are given the amount of sodium hydroxide (NaOH) in grams and the volume of the solution in milliliters.
To start, we need to convert the mass of NaOH to moles. The molar mass of NaOH is approximately 40 g/mol, so:
Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH
= 26 g / 40 g/mol
= 0.65 mol
Next, we need to convert the volume of the solution to liters since pH and pOH are usually expressed in terms of moles per liter. The volume is given as 150 mL, which is equal to 0.150 L.
Now, we can calculate the concentration of OH- ions in the solution. Since NaOH dissolves completely in water, it dissociates into one Na+ ion and one OH- ion. Thus, the concentration of OH- ions is equal to the concentration of NaOH.
Concentration of OH- ions = Number of moles of NaOH / Volume of solution
= 0.65 mol / 0.150 L
= 4.33 mol/L
Now that we have the OH- concentration, we can calculate the pOH. The pOH is defined as the negative logarithm (base 10) of the concentration of OH- ions:
pOH = -log[OH-]
= -log(4.33 mol/L)
≈ -0.638
The pH is related to pOH by the following equation:
pH + pOH = 14
Therefore, we can find the pH by subtracting the pOH from 14:
pH = 14 - pOH
= 14 - (-0.638)
≈ 14.638
So, the pH and pOH values for the solution are approximately 14.638 and -0.638, respectively.