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aev [14]
3 years ago
15

Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH

. What is the final pH? The Ka of acetic acid is 1.770 × 10-5
Chemistry
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

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Write the oxidation and reduction half reactions;
luda_lava [24]

Answer:

a)

Fe^{2+}⇒Fe^{3+}+e^-

Br_2+2e^-⇒2Br^-

b)

Mg⇒Mg^{2+}+2e^-

Cr^{3+}+e^-⇒Cr^{3+}

Explanation:

A)

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

Fe^{2+}⇒Fe^{3+}+e^-

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

Br_2+2e^-⇒2Br^-

b)

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

Mg⇒Mg^{2+}+2e^-

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

Cr^{3+}+e^-⇒Cr^{3+}

3 0
3 years ago
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Calculate the amount of oxygen gas collected by the displacement of water at 14◦C if the atmospheric pressure is 790 Torr and th
zhannawk [14.2K]

Answer : The amount of oxygen gas collected are, 0.217 mol

Explanation :

Using ideal gas equation :

PV=nRT

where,

P = pressure of gas = (790-12)torr=778torr=1.02atm     (1 atm = 760 torr)

V = volume of gas = 5 L

T = temperature of gas = 14^oC=273+14=287K

n = number of moles of gas = ?

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1.02atm)\times (5L)=n\times (0.0821L.atm/mol.K)\times (287K)

n=0.217mole

Thus, the amount of oxygen gas collected are, 0.217 mol

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