Calculate the pH of the solution that results when 27.38mL of 0.15M H2SO4(aq) is mixed with 18.50mL of 0.19M Ca(OH)2 (aq)

To calculate the pH of the solution, we need to find the concentration of the hydronium ion (H3O+) or the hydroxide ion (OH-) in the solution.

First, let's find the number of moles of H2SO4 and Ca(OH)2:

Moles of H2SO4 = Volume of H2SO4 (in L) * Concentration of H2SO4 (in mol/L)
= 27.38 mL * (1 L/1000 mL) * 0.15 mol/L
= 0.004107 mol

Moles of Ca(OH)2 = Volume of Ca(OH)2 (in L) * Concentration of Ca(OH)2 (in mol/L)
= 18.50 mL * (1 L/1000 mL) * 0.19 mol/L
= 0.003515 mol

Next, let's write the balanced equation for the reaction between H2SO4 and Ca(OH)2:

H2SO4(aq) + Ca(OH)2(aq) -> CaSO4(aq) + 2H2O(l)

From the balanced equation, we can see that 1 mole of H2SO4 reacts with 1 mole of Ca(OH)2 to produce 2 moles of water.

Since both reactants are strong acids and bases, they will completely dissociate in water. So we can say that the moles of H3O+ formed will equal the moles of H2SO4 used, and the moles of OH- formed will equal the moles of Ca(OH)2 used.

Therefore, the moles of H3O+ in the solution = 0.004107 mol
And the moles of OH- in the solution = 0.003515 mol

Since the reaction produces twice as many moles of water as H3O+, the moles of H3O+ remaining in the solution will be reduced by half:

Moles of H3O+ in solution = 0.004107 mol - (0.004107 mol / 2)
= 0.002054 mol

Now, let's calculate the concentration of H3O+ in the solution:

Concentration of H3O+ = Moles of H3O+ / Volume of Solution (in L)
= 0.002054 mol / (27.38 mL + 18.50 mL) * (1 L/1000 mL)
= 0.002054 mol / 0.04588 L
= 0.04477 mol/L

Finally, let's calculate the pH of the solution:

pH = -log[H3O+]
= -log(0.04477)
≈ 1.35

Therefore, the pH of the solution that results from mixing 27.38 mL of 0.15 M H2SO4(aq) with 18.50 mL of 0.19 M Ca(OH)2(aq) is approximately 1.35.

mL x M = millimoles.

27.38 x 0.15 = 4.197 mmoles H2SO4
18.5 x 0.19 = 3.515 mmoles Ca(OH)2
(Watch the significant figures. I have used too many assuming you did not omit zeros on the 0.15 and 0.19).

......H2SO4 + Ca(OH)2 ==> CaSO4 + 2H2O
init...4.107....3.515........0....0
change.-3.515...3.515....3.515.3.515
equil...0.592....0........3.515..3.515

So the ICE chart tells you that you have an excess of 0.592 mmoles H2SO4 in (27.38+18.5 mL) for a M = 0.592 moles/45.88 mL = ?M
Determine the pH of that solution
Remember H2SO4 has a k2 value of about 0.012.

0.012