Hydrolysis Reactions of Anions and Cations of Salts Laboratory Experiment Report

Question

Hydrolysis Reactions of Anions and Cations of Salts

Laboratory Experiment Report Form.

Please help me figure out the ph, oh, and h values in this table. Please no cursive handwriting.

Table 5.I: Indicator Colors Bromo- |Phenol Phenol Alizarin H (H1 (0H1 Solution Methyl Methyl Bromo- Red phtalein Yellow- Orange Red H,O red red blueres color less yelo (unboiled) red red ue red tolerless ReA iled 3.1 4.8 7 8.0 8.2 12 0.10M red red yellea color les yehow Na 3. 4.8 62 8.0 8.2 16.) NaC,H,o, o.I0 M ed red blue yello, Celor ess reo NH,CI yellow 6.2.6-ericss ZnCl red KA 3, 6.0 2 4 8.2 0. KAI(SO,), 010Mellow red blue ed ved red Na,CO

Explanation / Answer

1) The pH of H2O (unboiled) < 7

Reason: Unboiled water contains minerals, i.e. acidic impurities and due to this, the pH value is below 7.

pH = -Log[H+]

i.e. [H+] > 10-7 M

[H+][OH-] = 10-14, i.e. [OH-] < 10-7 M

2) The pH of H2O (boiled) = 7 (i.e. neutral)

Reason: Boiled water doesn't contain minerals, i.e. neither acidic impurities nor basic impurities and due to this, the pH value is almost 7.

pH = -Log[H+]

i.e. [H+] = 10-7 M

[H+][OH-] = 10-14, i.e. [OH-] = 10-7 M

3) The pH of 0.1 M NaCl = 7 (i.e. neutral)

Reason: NaCl results from the complete neutralization of a strong acid (HCl) with a strong base (NaOH) and due to this, the pH of NaCl solution is 7.

pH = -Log[H+]

i.e. [H+] = 10-7 M

[H+][OH-] = 10-14, i.e. [OH-] = 10-7 M

4) The pH of 0.1 M NaC2H3O2 = 7 + 1/2 (pKa + Log[NaC2H3O2])

= 7 + 1/2 (4.74 + Log0.1)

= 8.87

Reason: NaC2H3O2 results from the reaction of weak acid (C2H3OH) with strong base (NaOH) and due to this the pH value is above 7.

pH = -Log[H+]

i.e. [H+] = 10-8.87 M = 1.35*10-9 M

[H+][OH-] = 10-14, i.e. [OH-] = 7.41*10-6 M

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