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elena-14-01-66 [18.8K]
3 years ago
11

Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 1

2.8 g of hydrochloric acid is mixed with 7.0 g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Round your answer to significant digits.
Chemistry
1 answer:
mario62 [17]3 years ago
7 0

Answer:

Mass of HCl leftover = 6.4 g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NaOH + HCl —> NaCl + H₂O

Next, we shall determine the masses of NaOH and HCl that reacted from the balanced equation. This can be obtained as follow:

Mass of NaOH = 23 + 16 + 1

= 40 g/mol

Mass of NaOH from the balanced equation = 1 × 40 = 40 g

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Mass of HCl from the balanced equation = 1 × 36.5 = 36.5 g

SUMMARY:

From the balanced equation above,

40 g of NaOH reacted with 36.5 g of HCl.

Next, we shall determine the mass of HCl required to react with 7 g of NaOH. This can be obtained as follow:

From the balanced equation above,

40 g of NaOH reacted with 36.5 g of HCl.

Therefore, 7 g of NaOH will react with = (7 × 36.5)/40 = 6.4 g of HCl.

Thus, 6.4 g of HCl is required for the reaction.

Finally, we shall determine the leftover mass of HCl. This can be obtained as follow:

Mass of HCl given = 12.8 g

Mass of HCl that reacted = 6.4 g

Mass of HCl leftover =?

Mass of HCl leftover = (Mass of HCl given) – (Mass of HCl that reacted)

Mass of HCl leftover = 12.8 – 6.4

Mass of HCl leftover = 6.4 g

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Which of the following buffers will be most effective at pH 9.25? Group of answer choices a mixture of 1.0 M HC2H3O2 and 1.0 M N
ludmilkaskok [199]

Answer:

The most effective buffer at pH 9.25 will be  a mixture of 1.0 M NH3 and 1.0 M NH4Cl

Explanation:

Step 1: Data given

pH of a buffer = pKa + log ([A-]/[Ha])

a mixture of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (Ka for acetic acid = 1.8 x 10-5)

pH = -log( 1.8 * 10^-5) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 4.74

a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10)

pH = -log( 4.9 * 10^-10) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 9.30

a mixture of 1.0 M HCl and 1.0 M NaCl

The solution made from NaCl and HCl will NOT act as a buffer.

HCl is a strong acid while NaCl is salt of strong acid and strong base which do not from buffer solutions hence due to HCl PH is less than 7.

a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 10^-5)

Ka * Kb = 1*10^-14

Ka = 10^-14 / 1.76*10^-5

Ka = 5.68*10^-10

pH = -log( 5.68*10^-10) + log (1/1)

pH = -log( 5.68*10^-10)

pH = 9.25

The most effective buffer at pH 9.25 will be  a mixture of 1.0 M NH3 and 1.0 M NH4Cl

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3 years ago
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