What is the pH of a 50.0 mL solution containing 1.53 g of sodium sulfate. This is a diprotic base.
(Na2SO4) = grams/molar mass = approx 0.01 but that's just an estimate and that divided by 0.050 L = about 0.2 M
There are two ways to do this, the short way and the long way. The short way is not very exact but the long way is. The short way is to call Na2SO4 a salt of a strong base (NaOH) and a strong acid (H2SO4). Assuming that ALL of the H2SO4 ionizes completely into H^+ and SO4^2- ions then the pH is 7 for a neutral solution. since neither Na^+ nor SO4^2- is hydrolyzed.
However, H2SO4 actually has a K2 = about 0.012 and if you take that into account you get this from the hydrolysis of the sulfate.
..............SO4^2- + HOH ==> HSO4^- + OH^-
I..............0.2M...........................0.............0
C............-x.................................x.............x
E.........0.2-x................................x.............x
Kb for SO4^2- = (Kw/k2 for HSO4^-) = (x)(x)/(0.2-x)
Solve for x = (OH^-) and convert to pH. Remember that 0.2 is an estimate. This gives a pH of approx 7.7 and it requires that you solve the quadratic to get it; however, it is a little better answer than the 7 we assumed by the first method. If this is a beginning class I assume the prof wants the first solution. It it is an advanced class I think the prof will want the latter solution. kw is 0.012 in my text.
i dont understand where to start though
When I have shown you how to do this two ways, I don't understand what you don't understand. If you want to work it the short way you make the assumption I discussed and since neither Na^+ nor SO4^2- are hydrolyzed, then the pH is that of pure water which is 7.0.
If you think the prof wants you to work the problem the long way, then you start with that hydrolysis equation for SO4^2-, follow the ICE chart, and solve the equation I wrote for you. Substitute as shown and solve for OH^- and convert to pH.
To determine the pH of a solution, we need to know the concentration of the solute, which in this case is sodium sulfate (Na2SO4). However, the information provided only includes the mass of sodium sulfate (1.53 g) and the volume of the solution (50.0 mL).
To find the concentration of sodium sulfate, we need to convert the mass to moles, and then divide by the volume in liters (L). The molar mass of sodium sulfate (Na2SO4) is approximately 142.04 g/mol.
Step 1: Calculate the moles of Na2SO4:
moles = mass / molar mass
moles = 1.53 g / 142.04 g/mol
Step 2: Convert the volume from milliliters to liters:
volume = 50.0 mL / 1000 mL/L
Step 3: Calculate the concentration of Na2SO4:
concentration (in molarity) = moles / volume
Now that we have the concentration of sodium sulfate, we can determine the pH of the solution.
Since sodium sulfate is a diprotic base, it will produce two moles of hydroxide ions (OH-) for every mole of sodium sulfate that dissociates in water.
The formula for sodium sulfate is Na2SO4. In water, it dissociates into two sodium ions (2Na+), which are spectator ions and do not affect the pH, and one sulfate ion (SO4^2-), which reacts with water to produce two hydroxide ions (2OH-). Therefore, the number of hydroxide ions produced is twice the concentration of sodium sulfate.
Step 4: Calculate the concentration of hydroxide ions (OH-):
concentration of hydroxide ions = 2 * concentration of sodium sulfate
Now that we have the concentration of hydroxide ions, we can calculate the pOH. The pOH is the negative logarithm (base 10) of the hydroxide ion concentration.
Step 5: Calculate the pOH:
pOH = -log10(concentration of hydroxide ions)
Finally, we can calculate the pH. The pH is the negative logarithm (base 10) of the hydrogen ion concentration. Since water autoprotolysis provides an equal concentration of hydrogen ions (H+) and hydroxide ions (OH-), their concentrations will be equal and sum up to 14 (pH + pOH = 14).
Step 6: Calculate the pH:
pH = 14 - pOH
By following these steps and plugging in the appropriate values, you can calculate the pH of the given solution.