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ziro4ka [17]
3 years ago
10

A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a

calculate the ph of the solution upon the addition of 0.015 mol of naoh to the original buffer.
Chemistry
1 answer:
Mandarinka [93]3 years ago
6 0
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87
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Answer:

Q(total) = 283Kj

Explanation:

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Q = mcΔT for temperature change regions of the heating curve (single phase only)

Q = m∙ΔH for phase transition regions of the heating curve (2 phases in contact)

Solid (ice) => Melting Pt  => Q(s) = mcΔT = (90.5g)(0.50cal/g∙⁰C)(11⁰C) = 478 cal

Melting (s/l) => Liquid (water) =>   Q(s/l) = m∙ΔHᵪ = (90.5g)(80cal/g) = 7240 cal

Liquid (water) => Boiling Pt => Q(l) = mcΔT = (90.5g)(1.0cal/g∙⁰C)(100⁰C) = 9050 cal

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Gas (steam) => Steam @ 145⁰C => Q(g = mcΔT = (90.5g)(0.48cal/g∙⁰C)(45⁰C) = 2036 cal

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                                 = 67,674 cal x 4.184 j/cal = 283,148 joules = 283 Kj

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You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene. Upon complete combustion in oxygen, you collect 0.204 g o
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Answer:

The empirical formula is C7H5N3O6  

Explanation:

Step 1: Data given

Mass of sample = 0.150 grams

Mass of CO2 = 0.204 grams

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Mass of H2O = 0.030 grams

Molar mass H2O = 18.02 g/mol

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Step 2: Calculate moles CO2

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Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.00464 moles we have 0.00464 moles C

Step 4: Calculate mass C

Mass C = 0.00464 moles * 12.01 g/mol

Mass C = 0.0557 grams

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Moles H2O = 0.030 grams / 18.02 g/mol

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Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

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Step 7: Calculate mass H

Mass H = 0.00332 moles * 1.01 g/mol

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Step 9: Calculate moles N

Moles N = 0.02775 grams / 14.0 g/mol

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Step 11: Calculate moles O

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Step 11: Calculate mol ratio

We divide by the smallest amount of moles

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H: 0.00332 moles / 0.00198 moles = 1.66

N: 0.00198 moles / 0.00198 moles = 1

O: 0.00395 moles / 0.00198 moles = 2

For 1 mol N we have 2.33 moles C, 1.66 moles H and 2 moles O

OR

For 3 moles N we have 7 moles C, 5 moles H and 6 moles O

The empirical formula is C7H5N3O6  

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