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kari74 [83]
3 years ago
10

For each solution, calculate the initial and final pH after the addition of 0.010 mol of NaOH.

Chemistry
1 answer:
Murljashka [212]3 years ago
8 0

Answer:

A. 12.6

B. 4.05

C. 10.93

Explanation:

A. 0.010mol of NaOH in 250.0mL gives a concentration of 0.04M NaOH = 0.04M OH⁻

pOH = -log [OH⁻] = 1.398

pH = 14-pOH = <em>12.6</em>

<em></em>

B. The reaction of NaOH with HCHO₂ is:

NaOH + HCHO₂ → H₂O + CHO₂⁻ + Na⁺

Initial moles of CHO₂⁻ and HCHO₂ are:

CHO₂⁻ = 0.250L × (0.275mol/L) = 0.06875moles

HCHO₂ = 0.250L × (0.195mol/L) = 0.04875moles

After reaction:

CHO₂⁻ = 0.06875moles + 0.010mol = 0.07875mol

HCHO₂ = 0.04875moles - 0.010mol = 0.03875mol

Using H-H equation (pKa of this buffer: 3.74)

pH = 3.74 + log [CHO₂⁻] / [HCHO₂]

pH =  3.74 + log [0.07875mol] / [0.03875mol]

<em>pH = 4.05</em>

<em></em>

C. The reaction of NaOH with CH₃CH₂NH₃Cl is:

NaOH + CH₃CH₂NH₃Cl → H₂O + CH₃CH₂NH₂ + Cl⁻ + Na⁺

Initial moles of CH₃CH₂NH₃Cl and CH₃CH₂NH₂ are:

CH₃CH₂NH₃Cl = 0.250L × (0.235mol/L) = 0.05875moles

CH₃CH₂NH₂ = 0.250L × (0.255mol/L) = 0.06375moles

After reaction:

CH₃CH₂NH₃Cl = 0.05875moles - 0.010mol = 0.04875mol

CH₃CH₂NH₂ = 0.06375moles + 0.010mol = 0.07375mol

Using H-H equation (pKa of this buffer: 10.75)

pH = 10.75 + log [CH₃CH₂NH₂] / [CH₃CH₂NH₃Cl]

pH =  10.75 + log [0.07375mol] / [0.04875mol]

<em>pH = 10.93</em>

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How many mL of a 1.48 M calcium hydroxide solution are needed to neutralize 36.0 mL of a 1.63 M hydrochloric acid solution
Lelechka [254]

The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL

<h3>Balanced equation </h3>

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

  • The mole ratio of the acid, HCl (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 1

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
  • Volume of acid, HCl (Va) = 36 mL
  • Molarity of acid, HCl (Ma) = 1.63 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(1.63 × 36) / (1.48 × Vb) = 2

58.68 / (1.48 × Vb) = 2

Cross multiply

2 × 1.48 × Vb = 58.68

2.96 × Vb = 58.68

Divide both side by 2.96

Vb = 58.68 / 2.96

Vb = 19.8 mL

Learn more about titration:

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6 0
1 year ago
Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov
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Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field  = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3

Mass of the Astroturf = m

Density of the Astroturf = d = 187 oz/ft^3

d=\frac{m}{V}

m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz

1 oz = 0.0283495 kg

m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg

Weight of the Astroturf = W

W = mg

=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N

The weight of the Astroturf is 179,684.31 Newtons.

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