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melomori [17]
2 years ago
14

State and explain the relative change in the pH and in the buffer-component concentration ratio, [NaA]/[HA], for each of the fol

lowing additions:(d) Dissolve pure HA in the buffer
Chemistry
1 answer:
melisa1 [442]2 years ago
8 0

When pure HA is added to the buffer, the buffer component ratio and the pH decrease.

<h3>State and explain the relative change in the pH and in the buffer-component concentration ratio, [NaA]/[HA] for the dissolve of pure HA in the buffer.</h3>

When pure HA is added to the buffer, the buffer component ratio and the pH decrease. The added HA increases the concentrations of NA and HA. However, there is a greater relative increase in the concentration of HA. Hence, the ratio of [NaA]/[HA] decreases, causing the solution to become more acidic.

The capacity of a buffer to withstand pH change is measured. The concentration of the buffer's components namely, the acid and its conjugate base determine this ability. Greater buffer capacity is associated with higher buffer concentration.

To learn more about buffer-component, Visit:

brainly.com/question/9542245

#SPJ4

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Suppose that you have a 60.0% solution of NaOH. How many milliliters of water must be added to 30.0 mL of this solution to prepa
defon

Answer:

  • <u>21.4 ml (second choice)</u>

Explanation:

<u>1) Data:</u>

a) C₁ = 60.0% (initial solution)

b) V₁ = 30.0 ml (initial solution)

c) C₂ = 0% (pure water)

d) V₂ = ? (pure water)

e) C₃ = 35.0% (final concentration)

<u>2) Formula:</u>

  • C₁V₁ + C₂V₂ = C₃V₃
  • V₁ + V₂ = V₃ (assuming volume addtivity)

<u>3) Solution:</u>

<u />

a) Substitute values in the first formula:

  • 60.0% × 30 ml + 0 = 35% (30 ml + V₂)

b) Solve the equation (units are supressed just to manipulate the terms)

  • 18 = 10.5 + 0.35V₂

  • 0.35V₂ = 18 - 10.5 = 7.5

  • V₂ = 7.5 / 0.35 = 21.4 ml ← answer    
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