Estimate the pH of a 0.0015 M solution of sodium formate, HCOONa. The K, of form

Question

Estimate the pH of a 0.0015 M solution of sodium formate, HCOONa. The K, of formic acid (HCOOH) is 18101. Sodium formate completely dissociates to produce Na' and the formate ion (HCOO), which is the conjugate base of formic acid. (7.46) Calculate the pH of a solution prepared by mixing 100.0 mL of 0.0100 M acetic acid (CH,COOH) with 50.0 mL of 0.0100 M sodium acetate, the conjugate base of acetic acid. (K (acetic acid) 1.75-10. Sodium acetate completely dissociates in water to form Na' and the acetate ion (CH,COO). The Na' is a spectator ion. HINT: when you mix the solutions you are diluting the concentrations! (4.46)

Explanation / Answer

1) we know that

Kb = kw / Ka

so

Kb = 10-14 / 1.8 x 10-4

Kb =5.55 x 10-11

now

HCooNa is a weak base

we know that

for weak bases

[OH-] = sqrt ( Kb x C)

so

[OH-] =sqrt ( 5.55 x 10-11 x 0.0015)

[OH-] = 2.887 x 10-7

now

we know that

pOH = -log [OH-]

so

pOH = -log 2.887 x 10-7

pOH = 6.54

now

pH= 14 - pOH

pH = 14 - 6.54

pH = 7.46

so

the pH is 7.46

2)

we know that

moles = molarity x volume (L)

so

moles of CH3COOH = 0.01 x 0.1 = 10-3

moles of CH3COONa = 0.01 x 0.05 = 5 x 10-4

now

we know that

for a combination of weak acid and its conjugate base

pH = pKa + log [ conjugate base / acid ]

so

pH = -log Ka + log [CH3COONa / CH3COOH]

so

pH = -log 1.75 x 10-5 + log [ 5 x 10-4 / 10-3 ]

pH = 4.456

so

the pH is 4.46

3)

the reaction is

C6H5COONa + HCl ----> C6H5COOH + NaCl

now

moles of HCL   = 0.01 x 50 x 10-3 = 5 x 10-4

moles of C6H5COONa = 0.01 x 100 x 10-3 = 10-3

now

moles of C6H5COONa reacted = moles of HCl added = 5 x 10-4

now

moles of C6H5COONa remaining = 10-3 - 5 x 10-4 = 5 x 10-4

moles of C6H5COOH formed = moles of C6H5COONa reacted = 5 x 10-4

now

pH = -log Ka + log [C6H5COONa / C6H5COOH]

pH = -log 6.28 x 10-5 + log [ 5 x 10-4 / 5 x 10-4 ]

pH = 4.20

so

the pH is 4.20

4)

the reaction is given by

H2P04- + OH- ----> HP042- + H20

moles of OH- = moles of NaOH = 0.05 x 50 x 10-3 = 2.5 x 10-3

moles of H2P04- = 0.1 x 100 x 10-3 = 10 x 10-3

moles of H2p04- reacted = moles of OH- added = 2.5 x 10-3

now

moles of H2P04- remaining = 10 x 10-3 - 2.5 x 10-3 = 7.5 x 10-3

also

moles of HP042- formed = moles of H2P04- reacted = 2.5 x 10-3

now

we get

pH = pKa + log [ conjugate base / acid ]

pH = -log Ka + log [ HP042- / H2P04-]

pH = -log 6.3 x 10-8    + log [ 2.5 x 10-3 / 7.5 x 10-3 ]

pH =   6.723

so

the pH is 6.723

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