WRITE EQUATIONS FOR EACH REACTION AND SHOW ALL CALCULATIONS. Calculate the pH: 1

Question

WRITE EQUATIONS FOR EACH REACTION AND SHOW ALL CALCULATIONS. Calculate the pH: 1. 50 mL of 0.1 M HC H302 35 mL of 0.1 M NaOH 2. 50 mL of 0.1 M HC2H3O250 mL of 0.1 M NaOH 3. 50 mL of 0.1 M HC2H3O2+ 60 mL of 0.1 M NaOH 4. 10.00 mL of 0.25 M NH3 6.25 mL of 0.30 M HCI 5. 10.00 mL of 0.25 M NH3+ 25.00 mL of 0.10 M HCI 6. 25.00 mL of 0.25 M HF15.00 mL of 0.25 M KOH 7. 40.00 mL of 0.10 M CH3CH2COOH15.00 mL of 0.10 M NaOH 8. 50.00 mL of 0.10 M phosphoric acid 20.00 mL of 0.10 M NaOH. 9. A buffer is composed of 0.25 M NH4CI and 0.41 M NH3. What is its pH? 10. (a) 15.00 mL of 1 M CH3COOH are mixed with 25.0 mL of 0.85 M NaCH COO. What is the pH of the resulting mixture? (b) 1.00 mL of 0.1 M HCl are added to the above mixture. What is the resulting pH? K, CH3COOH/HC2H 02 1.8 x 10 K, HF 6.8 x 104 Kb NH,-1.8 x 10-5 K, CH3CH,COOH = 1.3 x 10.5 HsPOg Kal-7. 1x103, Ka2-63x108, ???-4.2x10-13

Explanation / Answer

3)
Given:
M(HC2H3O2) = 0.1 M
V(HC2H3O2) = 50 mL
M(NaOH) = 0.1 M
V(NaOH) = 60 mL


mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)
mol(HC2H3O2) = 0.1 M * 50 mL = 5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 60 mL = 6 mmol


We have:
mol(HC2H3O2) = 5 mmol
mol(NaOH) = 6 mmol

5 mmol of both will react

excess NaOH remaining = 1 mmol
Volume of Solution = 50 + 60 = 110 mL
[OH-] = 1 mmol/110 mL = 0.0091 M

use:
pOH = -log [OH-]
= -log (9.091*10^-3)
= 2.0414


use:
PH = 14 - pOH
= 14 - 2.0414
= 11.9586
Answer: 11.96

4)
Given:
M(HCl) = 0.3 M
V(HCl) = 6.25 mL
M(NH3) = 0.25 M
V(NH3) = 10 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.3 M * 6.25 mL = 1.875 mmol

mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 0.25 M * 10 mL = 2.5 mmol



We have:
mol(HCl) = 1.875 mmol
mol(NH3) = 2.5 mmol

1.875 mmol of both will react
excess NH3 remaining = 0.625 mmol
Volume of Solution = 6.25 + 10 = 16.25 mL
[NH3] = 0.625 mmol/16.25 mL = 0.0385 M
[NH4+] = 1.875 mmol/16.25 mL = 0.1154 M

They form basic buffer
base is NH3
conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {0.1154/3.846*10^-2}
= 5.222

use:
PH = 14 - pOH
= 14 - 5.2218
= 8.7782

Answer: 8.78

5)
Given:
M(HCl) = 0.1 M
V(HCl) = 25 mL
M(NH3) = 0.25 M
V(NH3) = 10 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.1 M * 25 mL = 2.5 mmol

mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 0.25 M * 10 mL = 2.5 mmol



We have:
mol(HCl) = 2.5 mmol
mol(NH3) = 2.5 mmol

2.5 mmol of both will react to form NH4+ and H2O
NH4+ here is strong acid
NH4+ formed = 2.5 mmol
Volume of Solution = 25 + 10 = 35 mL
Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10
concentration ofNH4+,c = 2.5 mmol/35 mL = 0.0714 M


NH4+      + H2O ----->     NH3   +   H+
7.143*10^-2                    0         0
7.143*10^-2-x                  x         x


Ka = [H+][NH3]/[NH4+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.556*10^-10)*7.143*10^-2) = 6.299*10^-6

since c is much greater than x, our assumption is correct
so, x = 6.299*10^-6 M



[H+] = x = 6.299*10^-6 M

use:
pH = -log [H+]
= -log (6.299*10^-6)
= 5.2007
Answer: 5.20

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