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lesya692 [45]
3 years ago
11

To determine the concentration of 20.00 mL of an unknown solution of a monoprotic acid, it is titrated with 0.1093 M sodium hydr

oxide. The initial reading on the buret of sodium hydroxide is 19.63 mL. The final reading is 41.63 mL. What is the concentration of the unknown acid?a) 0.09936 Mb) 1.000 Mc) 0.2275 Md) 0.1202 M
Chemistry
1 answer:
Elanso [62]3 years ago
7 0

Answer:

d) 0.1202 M

Explanation:

Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.

NaOH + HA → NaA + H₂O

The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:

22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol

The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.

The molar concentration of HA is:

2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M

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g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the
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7.49% of Mercury

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Let N₀ represent the original amount.

Let N represent the amount after 10 days.

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Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

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Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

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Take the anti log of 1.1255

N₀/N = anti log 1.1255

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N/N₀ = 0.0749 × 100

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Thus, 7.49% of Mercury will be remaining after 10 days

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