HOAc Ka = 1.8 * 10–5 H2CO3 Ka1 = 4.3 * 10–7 Ka2 = 5.6 * 10–11 Which of the following 0.01 M solutions has the highest pH?

A)
HOAc
B)
NaOAc
C)
Na2CO3
D)
H2CO3
E)
NaHCO3

1 Answer
Mar 10, 2018

(c) sf(Na_2CO_3)

Explanation:

(a)

This is a weak acid:

sf(HOAcrightleftharpoonsH^++OAc^-)

This is a long question so I will go straight to this expression for the pH of a weak acid, which can be derived from the ICE table:

sf(pH=1/2[pK_a-loga])

sf(a) is the concentration of the acid.

sf(pK_a=-logK_a=-log[1.8xx10^(-5)]=4.745)

:.sf(pH=1/2[4.75-(-2)])

sf(pH=color(red)(3.37))

(b)

This is the salt of a weak acid and a strong base so undergoes hydrolysis:

sf(OAc^(-)+H_2OrightleftharpoonsHOAc+OH^-)

sf(pK_a+pK_b=pK_w=14)

:.sf(pK_b=14-4.75=9.255)

We can use the corresponding expression for the pOH of a weak base:

sf(pOH=1/2[pK_b-logb])

sf(b) is the concentration of the base.

sf(pK_a+pK_b=14)

:.sf(pK_b=14-4.475=9.255)

:.sf(pOH=1/2[9.255-(-2)]=5.627)

sf(pH+pOH=14)

:.sf(pH=14-5.627=color(red)8.37)

(c)

Carbonate ions are weak bases:

sf(CO_3^(2-)+H_2OrightleftharpoonsHCO_3^(-)+OH^-)

sf(K_(b1)xxK_(a2)=K_w=10^(-14))

sf(K_(b1)=10^(-14)/(5.6xx10^(-11))=1.786xx10^(-4))

sf(pK_(b1)=-log[K_(b1)]=-log[1.786xx10^(-4)]=4.748)

sf(pOH=1/2[pK_(b1)-logb])

sf(pOH=1/2[4.78-(-2)]=3.39)

sf(pH=14-3.39=color(red)(10.61)

(d)

Carbonic acid is a weak acid:

sf(H_2CO_3rightleftharpoonsH^(+)+HCO_3^-)

sf(pK_(a1)=-log[K_(a1)]=-log[4.3xx10^(-7)]=6.37)

If sf(K_(a1) is greater than sf(K_(a2) by a factor of sf(10^3) then we can assume that virtually all the sf(H^+) ions come from the 1st dissociation.

:.sf(pH=1/2[6.37-(-2)]=color(red)4.19)

(e)

sf(HCO_3^(-)rightleftharpoonsH^(+)+CO_3^(2-))

sf(K_(a2)=5.6xx10^(-11))

sf(pK_(a2)=-log[K_(a2)]=-log[5.6xx10^(-11)]=10.25)

sf(pH=1/2[10.25-(-2)]

sf(pH=color(red)(6.13)