Answer:
a) 3.98 x 10^-10
Explanation:
Hello,
In this case, for the given pH, we can compute the concentration of hydronium by using the following formula:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
Hence, solving for the concentration of hydronium:
![[H^+]=10^{-pH}=10^{-9.40}\\](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-9.40%7D%5C%5C)
![[H^+]=3.98x10^{-10}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D3.98x10%5E%7B-10%7DM)
Therefore, answer is a) 3.98 x 10^-10
Best regards.
Answer: I think you did nothing wrong
Explanation: I think you did nothing wrong
Technically speaking, yes you can. Using a microscope though.
The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M
