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3 years ago

What is the pH of 0.000134 M solution of HCI?

Chemistry

2 answers:

emmainna [20.7K]3 years ago

7 0

Answer: The pH of 0.000134 M solution of HCl is 3.873.

Explanation:

When HCl dissociates in water then it will give hydrogen and chlorine ions. The reaction will be as follows.

$HCl \rightarrow H^{+} + Cl^{-}$

This means that 1 mole of HCl is giving 1 mole of hydrogen ions on dissociation.

Hence, pH of the given solution will be calculated as follows.

pH = $-log [H^{+}]$

= $-log (0.000134)$

= 3.873

Thus, we can conclude that the pH of 0.000134 M solution of HCl is 3.873.

mafiozo [28]3 years ago

5 0

Answer: The pH will be 3.87

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$HCl\rightarrow H^++Cl^{-}$

According to stoichiometry,

1 mole of $HCl$ gives 1 mole of $H^+$

Thus $0.000134$ moles of $HCl$ gives = $\frac{1}{1}\times 0.000134=0.0001342$ moles of $H^+$

Putting in the values:

$pH=-\log[0.000134]$

$pH=3.87$

Thus the pH will be 3.87

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A solution is prepared by mixing 250 mL of 1.00 M CH3COOH with 500 mL of 1.00 M NaCH3COO. What is the pH of this solution? (Ka f

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Answer:

A solution is prepared by mixing 250 mL of 1.00 M

CH3COOH with 500 mL of 1.00 M NaCH3COO.

What is the pH of this solution?

(Ka for CH3COOH = 1.8 × 10−5 )

Explanation:

This is a case of a neutralization reaction that takes place between acetic acid, CH 3 COOH , a weak acid, and sodium hydroxide, NaOH , a strong base.

The resulting solution pH, depends if the neutralization is complete or not. If not, that is, if the acid is not completely neutralized, a buffer solution containing acetic acid will be gotten, and its conjugate base, the acetate anion.

It's important to note that at complete neutralization, the pH of the solution will not equal 7 . Even if the weak acid is neutralized completely, the solution will be left with its conjugate base, this is the reason why the expectations of its pH is to be over 7 .

So, the balanced chemical equation for this reaction is the ionic equation:

CH 3 COOH (aq] + OH − (aq] → CH 3 COO − (aq] + H 2 O (l]

Notice that:

1 mole of acetic acid will react with: 1 mole of sodium hydroxide, shown here as hydroxide anions, OH − , to produce 1 mole of acetate anions:

CH 3 COO −

To determine how many moles of each you're adding , the molarities and volumes of the two solutions are used:

c = n / V ⇒ n = c ⋅ V

n acetic = 0.20 M ⋅ 25.00 ⋅ 10 − 3 L = 0.0050 moles CH3 COOH

and

n hydroxide = 0.10 M ⋅ 40.00 ⋅ 10 − 3 L = 0.0040 moles OH −

There are fewer moles of hydroxide anions, so the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution is:

n acetic remaining = 0.0050 − 0.0040 = 0.0010 moles

The reaction will also produce 0.0040 moles of acetate anions.

This is, then a buffer and the Henderson-Hasselbalch equation is applied to find its pH :

pH = p K a + log ( [ conjugate base ] / [ weak acid ] )

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base .

V total = V acetic + V hydroxide

V total = 25.00 mL + 40.00 mL = 65.00 mL

Thus the concentrations will be :

[ CH 3 COOH ] = 0.0010 moles / 65.00 ⋅ 10 − 3 L = 0.015385 M

and

[ CH 3 COO − ] = 0.0040 moles / 65 ⋅ 10 − 3 L = 0.061538 M

The p K a of acetic acid is equal to 4.75

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pH = 4.75 + log ( 0.061538 M / 0.015385 M )

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Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

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Putting values in equation 1, we get:

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By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

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As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago