Please can you assist me with the following problem I am totally muddled - I could swear I've never even heard of science!!!
H2SO4(aq) ---> 2H+(aq)+(SO4^2-)(aq)
A 0.21 g sample of sulfuric acid is dissolved completely in sufficient water to make 0.25 litre of final solution. Calculate the hydrogen ion concentration (in mol l^−1) in this solution. Give your answer in scientific notation to an appropriate number of significant figures. Remember to show the successive steps in the calculation, and to explain your reasoning.
What is the pH of the sulfuric acid solution (to the nearest whole number)?
Thanx but, I think I am a bit more confused.... please could you explain this in other words?
Calculate the moles of sulfuric acid.
molarity= molesAcid/volumeinliters
Hydrongen ion concnetration, assuming total ionization (in reality it is about 95 percent) = 2* molarity above? Why?
pH= -log of hydrogen ion concnetration.
Now all this assumes a lot about the mode of ionization, which mostly is not reality. In reality, sulfuric acid ionizes to H+ HSO4 easily, and ionizing the HSO4 into H+ and SO4-- is not as easy. I am assuming you are not figuring that in.
I think your instructor PROBABLY wants you to figure that in. I have worked it both ways, however, the pH, to the nearest whole number, is the same although to three places there is a difference. Here is how you do it the Loooong way.
H2SO4 ==> H^+ + HSO4^- k1= 100%
HSO4^- ==> H^+ + SO4^-2 k2.
The molarity of the H2SO4 initially is
0.21/98 mols and that in 0.25 L =
0.00857 mols/L. I may carry more significant figures than allowed but you can strike those at the end.
The first ionization will result in H^+ of 0.00857 mols/L.
The second ionization, which is not 100%, will be supressed by the H^+ from the first one and it has a k2 value.
k2 = [(H^+)(SO4^-2)/(HSO4^-)] = 1.2 x 10^-2. You should look us k2 in your text and use it. It may not be listed in your text as 0.012.
Now we plug the values into k2.
For (H^+) it is 0.00857 (from the first ionization) + y (from the second ionization). (SO4^-2) = y and (HSO4^-) = 0.00857-y). Multiply all of this out to obtain a quadratic. Solve the quadratic equation for y.
Then the total H^+ = sum of each ionization = 0.00857 + y from k2 and pH = - log(H^+).
Check my thinking. Post your work if you get stuck. Working it as Bob Pursley has suggested makes the assumption that the second ionization contributes the same as the first. That is true because k2 is relatively large. However, the quadratic solution allows for the small difference which you will see in the third decimal place if you go the long route. It is up to you to decide what you prof wants.
Sure, let's break it down step by step:
1. Calculate the moles of sulfuric acid:
To do this, you need to use the formula: moles = mass / molar mass.
In this case, you have a 0.21 g sample of sulfuric acid (H2SO4), and the molar mass of sulfuric acid is 98.09 g/mol.
So, moles = 0.21 g / 98.09 g/mol = 0.00214 mol.
2. Calculate the molarity of the solution:
Molarity (M) is defined as moles of solute divided by volume of solution in liters.
In this case, you have 0.00214 mol of sulfuric acid dissolved in 0.25 L of solution.
So, Molarity = 0.00214 mol / 0.25 L = 0.00857 mol/L.
3. Calculate the hydrogen ion concentration:
Since sulfuric acid (H2SO4) is a strong acid, it fully ionizes in water. So, 1 mole of sulfuric acid produces 2 moles of hydrogen ions (H+).
Therefore, the concentration of hydrogen ions in the solution is twice the molarity of sulfuric acid.
Hydrogen ion concentration = 2 * 0.00857 mol/L = 0.0171 mol/L.
4. Calculate the pH:
pH is a measure of the acidity or basicity of a substance. It is defined as the negative logarithm of the hydrogen ion concentration.
pH = -log[H+]
In this case, pH = -log(0.0171) = 1.767 (to three decimal places).
Rounding it to the nearest whole number, the pH of the sulfuric acid solution is 2.
Hope this helps and clarifies the steps involved in solving the problem!