Chemistry, 26.07.2019 18:30 Itskiyya

A50.0-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 added to the acid?

Answers: 1

Show answers

Answers

Answer from: elijah1090

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:

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

......(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 ${{\text{H}}^ + }$ 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 ?

2. Why is bromophenol blue used as an indicator for antacid titration?

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.

Answer from: Quest

in scientific notation, the value of the speed of light in a vacuum

will be written as 2.995 * 10^8. when this value is represented in the form of

a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

Answer from: Quest

answer///carbon;

Another question on Chemistry

Chemistry, 21.06.2019 18:30

Which orbitals form a pi bond? a.the s orbital and three p orbitals b.the s orbital and two p orbitals c.overlapping p orbitals d.overlapping hybrid orbitals

Answers: 2

Answer

Chemistry, 22.06.2019 05:30

Astudent carefully transfers 30 g of water and 30 g of alcohol in a glass tube, forming two layers and filling the tube completely. after sealing the tube, the student mixes the solutions, and notices a bubble that forms in the tube. what is the mass of the contents in the glass tube after mixing?

Answers: 2

Answer

Chemistry, 22.06.2019 10:00

What is the atomic mass of an atom that has 6 protons, 6 neutrons, and 6 electrons? a) 6 b) 8 c) + 1 d) 12 e) 18

Answers: 1

Answer

Chemistry, 22.06.2019 11:30

What is the main reason why some developing countries fear the increase the free trade policies around the world?

Answers: 2

Answer

You know the right answer?

A50.0-ml sample of 0.50 m hcl is titrated with 0.50 m naoh. what is the ph of the solution after 28....