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bezimeni [28]
3 years ago
12

Silver chloride is virtually insoluble in water so that the reaction appears to go to completion. How many grams of solid NaCl m

ust be added to 25.0 mL of 0.366 M AgNO3 solution to completely precipitate the silver
Chemistry
1 answer:
masya89 [10]3 years ago
4 0

Answer:

0.535 g

Explanation:

The reaction that takes place is:

  • NaCl + AgNO₃ → AgCl + NaNO₃

First we <u>calculate how many AgNO₃ moles are there in 25.0 mL of a 0.366 M solution</u>, using the <em>definition of molarity</em>:

  • Molarity = moles / liters
  • moles = Molarity * liters

<em>Converting 25.0 mL to L </em>⇒ 25.0 / 1000 = 0.025 L

  • moles = 0.366 M * 0.025 L = 0.00915 mol AgNO₃

Then we <u>convert AgNO₃ moles into NaCl moles</u>:

  • 0.00915 mol AgNO₃ * \frac{1molNaCl}{1molAgNO_3} = 0.00915 mol NaCl

Finally we<u> convert NaCl moles into grams</u>, using its <em>molar mass</em>:

  • 0.00915 mol NaCl * 58.44 g/mol = 0.535 g
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What property of liquid oxygen makes it especially difficult and potentially harmful to work with at home?
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A 50.0-ml sample of 0.50 m hcl is titrated with 0.50 m naoh. what is the ph of the solution after 28.0 ml of naoh have been adde
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The pH of the solution after addition of 28 mL of NaOH is added to HCl is \boxed{{\text{0}}{\text{.85}}} .

Further Explanation:

The proportion of substance in the mixture is called concentration. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

{\text{Molarity of solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume }}\left({\text{L}} \riht){\text{ of solution}}}}          

                             ......(1)        

                         

Rearrange equation (1) to calculate the moles of solute.

{\text{Moles}}\;{\text{of}}\;{\text{solute}}=\left( {{\text{Molarity of solution}}}\right)\left({{\text{Volume of solution}}}\right)       ......(2)

Substitute 0.50 M for the molarity of solution and 50 mL for the volume of solution in equation (2) to calculate the moles of HCl.

\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{HCl}}&= \left({{\text{0}}{\text{.50 M}}}\right)\left( {{\text{50 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}} \right)\\&= 0.02{\text{5 mol}}\\\end{aligned}

Substitute 0.50 M for the molarity of solution and 28 mL for the volume of solution in equation (2) to calculate the moles of NaOH.

\begin{aligned}{\text{Moles}}\;{\text{of}}\;{\text{NaOH}}&=\left( {{\text{0}}{\text{.50 M}}} \right)\left( {{\text{28 mL}}} \right)\left( {\frac{{{\text{1}}{{\text{0}}^{ - 3}}{\text{ L}}}}{{{\text{1 mL}}}}}\right)\\&= 0.014{\text{ mol}}\\\end{aligned}

The reaction between HCl and NaOH occurs as follows:

{\text{NaOH}} + {\text{HCl}} \to {\text{NaCl}} + {{\text{H}}_2}{\text{O}}

The balanced chemical reaction indicates that one mole of NaOH reacts with one mole of HCl. So the amount of remaining HCl can be calculated as follows:

\begin{aligned}{\text{Amount of HCl remaining}}&= 0.02{\text{5 mol}} - 0.01{\text{4 mol}}\\&= {\text{0}}{\text{.011 mol}} \\\end{aligned}

The volume after the addition of NaOH can be calculated as follows:

\begin{aligned}{\text{Volume of solution}} &= {\text{50 mL}} + {\text{28 mL}}\\&= {\text{78 mL}}\\\end{aligned}

Substitute 0.011 mol for the amount of solute and 78 mL for the volume of solution in equation (1) to calculate the molarity of new HCl solution.

\begin{aligned}{\text{Molarity of new HCl solution}}&= \left({{\text{0}}{\text{.011 mol}}} \right)\left( {\frac{1}{{{\text{78 mL}}}}}\right)\left( {\frac{{{\text{1 mL}}}}{{{{10}^{ - 3}}\;{\text{L}}}}} \right)\\&= 0.1410{\text{2 M}}\\&\approx {\text{0}}{\text{.141 M}}\\\end{aligned}

pH:

The acidic strength of an acid can be determined by pH value. The negative logarithm of hydronium ion concentration is defined as pH of the solution. Lower the pH value of an acid, the stronger will be the acid. Acidic solutions are likely to have pH less than 7. Basic or alkaline solutions have pH more than 7. Neutral solutions have pH equal to 7.

The formula to calculate pH of an acid is as follows:

{\text{pH}}=- {\text{log}}\left[ {{{\text{H}}^ + }}\right]     ......(3)

Here,

\left[{{{\text{H}}^ + }}\right] is hydrogen ion concentration.

HCl is a strong acid so it dissociates completely. So the concentration of   also becomes 0.141 M.

Substitute 0.141 M for \left[{{{\text{H}}^ + }}\right] in equation (3).

\begin{aligned}{\text{pH}}&= - {\text{log}}\left({0.141} \right)\\&=0.85\\\end{aligned}

So the pH of the solution is 0.85.

Learn more:

1. Which indicator is best for titration between HI and  ? brainly.com/question/9236274

2. Why is bromophenol blue used as an indicator for antacid titration? brainly.com/question/9187859

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Acid-base titrations

Keywords: molarity, pH, HCl, NaOH, 0.85, 0.141 M, moles of HCl, moles of NaOH, 50 mL, 0.50 M, 28 mL, 0.025 mol, 0.014 mol, 0.011 mol, 78 mL.

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