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3 years ago

Calculate the change in ph when hcl is added to a buffer solutio that is nh3 and nh4

1 answer:

8 0

1.) Calculate the change in pH when 4.00 ml of 0.100 M HCl(aq) is added to 0.100 mL of a buffer solution that is 0.100 M in NH3 (aq) and 0.1000 M in NH4Cl(aq).
2.) Calculate the change in pH when 4.00 mL of 0.100 M in NaOH(aq) is added to the original buffer solution.

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The word epi in Greek means "upon" or "over." Based on this definition, what body part likely contains epithelial tissue? In you

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Answer:

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Explanation:

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3 years ago

Write the acid-base reaction that occurred in the titration of potassium hydrogen phthalate(KHP), KHC8H4O4with sodium hydroxide,

ExtremeBDS [4]

Answer:

KHC₈H₄O₄ + NaOH → KC₈H₄O₄⁻ + Na⁺ + H₂O

Explanation:

Potassium hydrogen phthalate is an acidic salt that is used as a primary standard for acid-base titrations.

In the titration of KHP with NaOH, the salt loses a H⁺ producing water and Na⁺. The reaction is:

KHC₈H₄O₄ + NaOH → KC₈H₄O₄⁻ + Na⁺ + H₂O

I hope it helps!

8 0

3 years ago

The chemical formula of aspirin is C9H8O4. What is the mass of 0.40 mol<br> of aspirin?

dybincka [34]

Answer:

$\boxed {\boxed {\sf About \ 72 \ grams }}$

Explanation:

To convert from moles to grams, we must use the molar mass (also known as the gram formula mass).

First, look up the molar masses of the elements in the formula.

C: 12.011 g/mol
H: 1.008 g/mol
O: 15.999 g/mol

Next, multiply by the subscript, because it tells us the number of atoms of each element in the formula.

C₉: 4(12.011 g/mol)= 108.099 g/mol
H₈: 8(1.008 g/mol)= 8.064 g/mol
O₄: 4(15.999 g/mol)= 63.996 g/mol

Add the values.

108.099 + 8.064+63.996=180.159 g/mol

Use this molar mass as a ratio.

$\frac {180.159 \ g \ C_9H_8O_4}{1 \ mol \ C_9H_8O_4}$

Multiply by the given number of moles, 0.40

$0.40 \ mol \ C_9H_8O_4 *\frac {180.159 \ g \ C_9H_8O_4}{1 \ mol \ C_9H_8O_4}$

The units moles of aspirin cancel.

$0.40 *\frac {180.159 \ g \ C_9H_8O_4}{1 }$

$72.0636 \ g \ C_9H_8O_4$

The original number of moles has 2 sig figs (4 and 0), so answer must have the same. For the number we calculated, it is the ones place. The 0 in the tenth place tells us to leave the 2.

$72 \ g$

5 0

3 years ago

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

hram777 [196]

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.

4 0

4 years ago

Read 2 more answers

In which type of chemical reaction do two or more reactants combine to form one product, only?

xenn [34]

In decomposition, two or more reactant combine to form one product only.

8 0

3 years ago