200 ml of 0.1 M sodium hydroxide is mixed with 100 ml of 0.2 M ethanoic acid. What is the pH of the resulting solution?

2 Answers
Dec 2, 2016

Answer:

#sf(pH=8.8)#

Explanation:

The alkali neutralises the acid:

#sf(CH_3COOH+NaOHrarrCH_3COO^(-)Na^(+)+H_2O)#

The moles reacting (n) of each species is given by:

#sf(n_(OH^-)=cxxv=0.1xx200/1000=0.02)#

#sf(n_(CH_3COOH)=cxxv=0.2xx100/1000=0.02)#

This shows that they are added in the same mole ratio as defined by the equation.

We can, therefore, say that the no. moles of #sf(CH_3COO^(-))# formed = 0.02.

A salt formed from a weak acid and a strong base is slightly alkaline due to hydrolysis:

#sf(CH_3COO^(-)+H_2OrightleftharpoonsCH_3COOH+OH^(-))#

From an ICE table we can use this expression which applies to a weak base:

#sf(pOH=1/2(pK_b-logb)#

Where #sf(b)# is the concentration of the base which, in this case, is #sf(CH_3COO^(-))#.

The total volume is now 200 ml + 100 ml = 300 ml = 0.3 L

#sf([CH_3COO^(-)]=n/v=0.02/0.3=0.0666color(white)(x)"mol/l")#

You need to look up the #sf(pK_a)# value for ethanoic acid which = 4.76.

To get #sf(pK_b)# you use:

#sf(pK_(a)+ pK_(b)=14)#

#:.##sf(pK_b=14-pK_a=14-4.76=9.24)#

Putting in the numbers:

#sf(pOH=1/2[9.24-log(0.0666)]#

#sf(pOH=1/2[9.24-(-1.176)]=5.208)#

#sf(pH+pOH=14)#

#:.##sf(pH=14-pOH=14-5.208=8.8)#

As expected the solution is slightly alkaline, even though the acid and base have been added in the same molar ratio.

Dec 2, 2016

Answer:

#"pH" = 8.8#

Explanation:

This is really a two-part question:

  1. What is in the mixture after the reaction is complete?
  2. What is the #"pH"# of the mixture?

What's in the mixture?

I like to use an ICE table to keep track of the stoichiometry calculations.

#color(white)(mmmmll)"HAc" +color(white)(l) "OH" → "H"_2"O" + "Ac"^"-"#; #K_text(a) = 1.76 × 10^"-5"#
#"I/mol:"color(white)(ml)0.020color(white)(ml)0.020color(white)(mmmmmll)0#
#"Cmol:"color(white)(m)"-0.020"color(white)(m)"-0.020"color(white)(mmmm)"+0.020"#
#"E/mol:"color(white)(mm)0color(white)(mmml)0color(white)(mmmmmll)0.020#

#"Initial moles HAc" = 0.100 color(red)(cancel(color(black)("L"))) × "0.2 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.020 mol"#

#"Initial moles OH"^"-"= 0.200 color(red)(cancel(color(black)("L"))) × "0.1 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.020 mol"#

We have 300 mL of a solution that contains 0.020 mol of #"Ac"^"-"#.

2. What's the #"pH"# of the solution?

#"HAc"# is a weak acid, and #"Ac"^"-"# is its conjugate base.

The solution will be basic.

#K_"b" = K_"w"/K_"a" = (1.00 × 10^"-14")/(1.76 × 10^"-5") = 5.68 × 10^"-10"#

Now, we set up another ICE table (with molarities) to calculate the #"pH"# at equilibrium.

#["Ac"^"-"] = "0.020 mol"/"0.300 L" = "0.067 mol/L"#

#color(white)(XXXXXX)"Ac"^"-" + "H"_2"O" ⇌ "HAc" + "HO"^"-"#
#"I/mol:"color(white)(Xm) 0.067color(white)(XXXXXXll) 0color(white)(mmll)0#
#"C/mol:"color(white)(mml) "-"xcolor(white)(XXXXXXlll) "+"xcolor(white)(mll)"+"x#
#"E/mol:"color(white)(Xl)"0.067-"xcolor(white)(XXXXXlll) xcolor(white)(mmll)x#

#K_b = (["HAc"]["HO"^"-"])/(["Ac"^"-"]) = x^2/(0.067-x) = 5.68 × 10^"-10"#

Check if #x ≪ 0.067#.

#0.067/(2.10 × 10^"-10") = 3.2 × 10^8 > 400#; ∴ we can ignore #x#.

#x^2/0.067 = 5.68 × 10^"-10"#

#x^2 = 0.067 × 5.68 × 10^"-10" = 3.8 × 10^"-11"#

#x = 6.2 × 10^"-6"#

#["HO"^"-"] = 6.2 × 10^"-6" color(white)(l)"mol/L"#

#"pOH" = "-log"(6.2 × 10^"-6") = 5.21#

#"pH" = "p"K_"w" - "pOH" = "14.00 - 5.21" = 8.8#