Respuesta :
The change in ph when 9.00 ml of 0.100 m hcl(aq) is added to 100.0 ml of a buffer solution that is 0.100 m in nh3(aq) and 0.100 m in nh4cl(aq). consult
1)
Step 1: Calculate initial pH
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 {0.1/0.1}
= 4.745
use:
PH = 14 - pOH
= 14 - 4.7447
= 9.2553
Step 2: Calculate pH after adding HCl
mol of HCl added = 0.1M *9.0 mL = 0.9 mmol
NH3 will react with H+ to form NH4+
Before Reaction:
mol of NH3 = 0.1 M *100.0 mL
mol of NH3 = 10 mmol
mol of NH4+ = 0.1 M *100.0 mL
mol of NH4+ = 10 mmol
after reaction,
mol of NH3 = mol present initially - mol added
mmol of NH3 = (10 - 0.9) mmol
mol of NH3 = 9.1 mmol
mol of NH4+ = mol present initially + mol added
mol of NH4+ = (10 + 0.9) mmol
mol of NH4+ = 10.9 mmol
since volume is both in numerator and denominator, we can use mol instead of concentration
Kb = 1.8*10^-5
pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745
After the reaction, the following equation holds true: mol of NH3 = mol present initially - mol added mmol of NH3 = (10 - 0.9) mmol mol of NH3 = 9.1 mmol mol of NH4+ = mol present initially + mol added mol of NH4+ = (10 + 0.9) Since volume appears in both the numerator and denominator of the equation, we can use mol in place of concentration: mmol mol of NH4+ = 10.9 mmol
Kb = 1.8*10-5 pKb = -log(Kb) = -log(1.8*10-5) = 4.745
Employ: pOH = pKb + log [conjugate acid]/[base] = 4.745 + log [10.9/9.1] = 4.823
employ: PH = 14 - pOH = 14 - 4.8231 = 9.1769
Step 3: Subtract the final pH from the initial pH to arrive at a change delta pH of -0.0784 (9.1769 - 9.2553).
Incorrect: -0.0784
2) The reaction between NH4+ and OH- will produce NH3 when mol of NaOH supplied is 0.1M *9.0 mL = 0.9 mmol.
NH3 = 0.1 M * 100.0 mL mol of NH3 = 10 mmol NH4+ = 0.1 M * 100.0 mL mol of NH4+ = 10 mmol
After reaction, the equation is: Mol of NH3 = Mol originally present + Mol of NH3 added = (10 + 0.9). mmol mol of NH3 equals 10.9 mmol mol of NH4+ equals mol initially present minus mol added mol of NH4+ = (10 - 0.9) Since volume is in both the numerator and the denominator of mmol mol of NH4+ = 9.1 mmol, we can substitute mol for concentration.
Kb = 1.8*10-5 pKb = -log(Kb) = -log(1.8*10-5) = 4.745
Make use of: pOH = pKb + log [conjugate acid]/[base] = 4.745 + log [9.1/10.9] = 4.666
employ: PH = 14 - pOH = 14 - 4.6663 = 9.3337
delta pH = final pH - initial pH
= 9.3337 - 9.2553
= 0.0784
Answer: 0.0784
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