chem questions

[nooshnoosh95]

What is the resulting pH if 25mL of 0.1 Ba(OH)2  solution is added to 25mL of 0.1 HCl solution?
could you show me how to work this out please

[thushan]

What is the resulting pH if 25mL of 0.1 Ba(OH)2  solution is added to 25mL of 0.1 HCl solution?
could you show me how to work this out please

OK, to work out pH, we need to work out either [H+] or [OH-]. So we need either n(H+) or n(OH-) and the volume of solution (which is 25 + 25 = 50 mL).

n(OH-)added = 2n(Ba(OH)2) = 2 x 0.1 x 0.025 = 0.0050 mol
n(H+)added = n(HCl) = 0.1 x 0.025 = 0.0025 mol

OH- is in excess, so n(OH-)final = 0.0050 - 0.0025 = 0.0025 mol

[OH-] = 0.0025/0.050 = 0.050 M
pOH = -log (base 10) [OH-] = 1.30
Therefore, pH = 14 - pOH = 12.70

[nooshnoosh95]

makes sense

thank you xx

[pi]

What is the resulting pH if 25mL of 0.1 Ba(OH)2  solution is added to 25mL of 0.1 HCl solution?
could you show me how to work this out please

OK, to work out pH, we need to work out either [H+] or [OH-]. So we need either n(H+) or n(OH-) and the volume of solution (which is 25 + 25 = 50 mL).

n(OH-)added = 2n(Ba(OH)2) = 2 x 0.1 x 0.025 = 0.0050 mol
n(H+)added = n(HCl) = 0.1 x 0.025 = 0.0025 mol

OH- is in excess, so n(OH-)final = 0.0050 - 0.0025 = 0.0025 mol

[OH-] = 0.0025/0.050 = 0.050 M

Alternate path from there:

pH = 14 + log[OH-] = 14 + log(0.050) = 1 x 10^1 (one sig fig)

[charmanderp]

^that's actually two sig figs?

[pi]

^that's actually two sig figs?

Woops! Fixed