Suppose that 1.2 g of NaOH is dissolved in 500mL of a aquous buffer that is 0.400 M in NaC2H3O2 and 0.800 M in HC2H3O2.
Calculate the resulting pH and the change in pH.
Please show work so I can figure it out...I've been having a hard time with this!
Thanks!
2006-11-06 05:18:44 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
As in 500 ml of buffer has concentration of 0.800 M HC2H3O2(acetic acid), which is equivalent to 0.4 moles of acetic acid ,so it will be neutralised by 1.2 g i.e .. 1.2/40 =0.03 moles of NaOH.
so now excess of acetic acid = 0.4 - 0.3 =0.1 mole in 500 ml of buffer. so now conc. of acetic acid is = 0.2 M
Now u can apply the formula,
pH = pKa + log [CH3COONa] / [CH3COOH]; [CH3COONa] =0.4M and [CH3COOH] = 0.2 M and pKa of acetic acid you can get from table.
by this way u can get pH. and initial pH can also be calculated by the same formula just replace the [CH3COOH] = 0.8M.
so change = initial pH - final pH.
2006-11-06 06:50:00 · answer #1 · answered by sandeep t 1 · 0⤊ 0⤋
pH = pKa + log [A-]/[HA]
2006-11-06 05:21:24 · answer #2 · answered by shiara_blade 6 · 0⤊ 0⤋