What is the pH of a solution made by mixing 0.10 L of 6.0 M NaOH with 0.90 L of 1.0 M H2SO4?

First of all this is a limiting reactant problem. Lets first find which is which.

Overall reaction:

2NaOH + H2SO4 --> 2H2O + Na2SO4

Now lets find moles of each reactant.

0.10L x (6.0mol NaOH/1L soln) = 0.6 mol NaOH

0.90L x (1.0mol H2SO4/1L soln) = 0.9 mol H2SO4

there needs to be 2 moles of NaOH for every mole of H2SO4 (according to the balanced equation). Therefore, NaOH is clearly the limiting reagent.

Now the stoichiometry.

0.6 mol NaOH x (1 mol H2SO4/2 mol NaOH) = 0.3 mol H2SO4 used in reaction.

mol H2SO4 = 0.9 mol - 0.3 mol = 0.6 mol H2SO4

Keep following with me here.... Dont get lost in that dark jungle now.

Hydrogen ion concentration = [H+]

[H+] = (0.6molH2SO4/1.0L soln) x (2mol H+/1mol H2SO4) = 1.2mol H+/1L soln

Since we have 1L of soln we can continue on with 1.2moles of H+.

the p in pH simply means -log[H+ concentration]

therefore pH = -log [1.2] = -.08 (which is extremely acidic).

if you hadd 600.0 mL then the volume changes to 1.6L. The [H+] will then be 0.75 M H+

pH = -log [0.75] = .12 (which is still really acidic)

I'm just continuing from where Philip R left off. This problem is easier if you convert the molarity to normality. A 0.60M sol'n of NaOH is 0.60N since there is only one equivalent of hydroxide. A 0.90M H2SO4 sol'n is 1.8N, 2eq per molecule. That leaves us with a 1.2N H2SO4 sol'n, divide by # of equivalents to get a 0.60M sol'n.

pH = -log [H30+] or pH = -log[0.60] which is 0.22

Anyway, the 2nd part: we have a 1L of a 0.60M sol'n, and 0.60 mol and a vol. of 1L. If we add 600.0 mL of water we then have 0.60 mol in 1.60L of water or 0.38M sol'n which pH would be

pH = -log[0.38] or 0.42

one million.5L / one thousand = 0.0015ml * .35M = 5.25 x 10^-4 mmol of 7bcfbacd98cfa6ce29d0ad821b14f9Cl 2L / one thousand = 0.002ml * .1M = 2.0 x 10^-4 mmol of NaO7bcfbacd98cfa6ce29d0ad821b14f9 proceed to be 7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9 = (5.25 x 10^-4) - (2.0x 10^-4) = 3.25 x 10^-4 moles 7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9 (aq) finished ml = 0.0015ml H+ 0.002ml = 0.0035ml [7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9] = (3.25 x 10^-4 moles 7bcfbacd98cfa6ce29d0ad821b14f97bcfbacd98cfa6ce29d0ad821b14f9) / (0.0035ml) = 0.0929 p7bcfbacd98cfa6ce29d0ad821b14f9 = -Log(0.0929) = one million.03