Help please! 1)Explain the reasons for the differences in pH for 0.1 M HCl and 0
ID: 1019988 • Letter: H
Question
Help please!
1)Explain the reasons for the differences in pH for 0.1 M HCl and 0.001 M HCl. ?
2)Explain the reasons for the differences in pH for 0.1 M HCl and 0.1 M HC 2H 3O 2 ?
3)Using the measured pH of the 0.1 M acetic acid solution determine the value of the ?acid dissociation constant, K a for acetic acid.?Using the measured pH for 0.1 M sodium acetate solution, also determine the K a for acetic acid. Compare these two values. ??
4). Explain how NaH2PO4 ; Na2HPO4 ; Na3PO4 ; H3PO4 are related to each other
pH (Calculated) pH (paper) #1 Solution pH 1 0.1 M HCI 2 0.01 M HCI 3 0.001 M HCl 4 0.1 M Nacl 50.1M HC2H302 6 0.1 M NaOH 7|deionized water 8Tap water 9 0.1 M NH3 100.1 M NH^Cl 110.1M (NH)2C20 120.1 M NaC,H20 130.1M NaH2PO,4 14 0.1 M Na,HPO 15 0.1 M Na,PO 160.1M H3PO (pH meter) 1.64 2.48 3.00 5.25 3.02 12.55 9.3 6.7 11.9 6.09 7.79 7.29 4.75 9.25 12.25 2.78 2 4 6.5 7 10 3 6 8 13 2Explanation / Answer
HCl is strong acid and ionizes completely
HCl ------> H+ Cl-
So 0.1 M HCl means 0.1M H+ and pH= -log [H+] =-log(0.1)= 1
for 0.001M HCl, pH= -log (0.001)= 3
2. Acetic acid is weak acid and does not ionize completely while HCl is strong acid and ionizes completely and that is reflected in the pH values.
3. CH3COOH----->CH3COO- +H+
pH= 3.02, [H+] =10(-3.02)= 0.000955 =[CH3COO-]
Ka= [H+] [CH3COO-]/ [CH3COOH] =0.000955*0.000955/0.1=9.1*10-6
CH3COONa is strong salt and ionizes completely
CH3COO-+ H2O--------CH3COOH + OH-
pH= 7.29, pOH= 14-7.29=6.71
[OH-] =10^(-6.71)=1.95*10-7
Kb=( 1.95*10-7*1.95*10-7/0.1 =3.8*10-13
Ka= 10-14/3.8*10-13= 0.0263
NaH2PO4 ; Na2HPO4 ; Na3PO4 ; H3PO4 are related to each other in thier pH as shown in the table. H3PO4 reacts with NaOH to give nAH2PO4, Na2HP4, Na3PO4. This leads to basicity.
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