the hydrogen ion concentration in a mixture of 10ml 0.1 M H2SO4 AND 10ml 0.1M KOH solution in water is?
To find the hydrogen ion concentration in the mixture of H2SO4 and KOH solutions, we need to calculate the total amount of hydrogen ions produced or consumed in the reaction between these two solutions.
The balanced chemical equation for the reaction between H2SO4 (sulfuric acid) and KOH (potassium hydroxide) is:
H2SO4 + 2KOH → K2SO4 + 2H2O
From the equation, we can see that one mole of sulfuric acid (H2SO4) reacts with two moles of potassium hydroxide (KOH) to produce two moles of water (H2O).
Given that both solutions are 0.1 M (moles per liter), we need to calculate the number of moles of each component in 10 mL (0.01 liters) of solution.
For H2SO4:
Molarity (M) = moles / volume (L)
0.1 M = moles / 0.01 L
moles = 0.1 M * 0.01 L
moles = 0.001 moles
For KOH:
Molarity (M) = moles / volume (L)
0.1 M = moles / 0.01 L
moles = 0.1 M * 0.01 L
moles = 0.001 moles
Since the stoichiometry of H2SO4 and KOH is 1:2, the mole ratio is 1:2. Therefore, the moles of H2SO4 consumed will be equal to half the number of moles of KOH.
So, moles of H2SO4 consumed = 0.001 moles / 2 = 0.0005 moles.
Now, let's calculate the concentration of hydrogen ions (H+) in the mixture.
Since the equation shows that two moles of water are produced for every mole of hydrogen ions, we can say that the number of moles of hydrogen ions is equal to the number of moles of water produced.
moles of H+ = 0.0005 moles
To find the concentration, we divide the moles of hydrogen ions by the total volume in liters.
Total volume = 10 mL + 10 mL = 20 mL = 0.02 L
Hydrogen ion concentration = moles of H+ / volume (L)
= 0.0005 moles / 0.02 L
= 0.025 M
Therefore, the hydrogen ion concentration in the mixture is 0.025 M.