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s2008m [1.1K]
3 years ago
5

A weak acid, like Hydrofluoric acid, can be dangerous if _____. Select all that apply. concentrated

Chemistry
1 answer:
Whitepunk [10]3 years ago
5 0

Answer:

Used without adult supervision

Used without goggles

Used without gloves

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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
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
3 years ago
Read 2 more answers
Answer the following questions: On a 10-fold dilution of a weak acid, the pH will ______________ . On a 10-fold dilution of a we
Scorpion4ik [409]

Answer:

i) increase

ii) decrease

iii) remain the same

iv) No, because it dissociates completely.

Explanation:

On a 10-fold dilution of a weak acid, the pH will increase because the concentration of hydrogen ions will decrease thereby increasing the pH to close to that of water.

On a 10-fold dilution of a weak base, the pH will decrease due to the removal of hydroxide ions from the solution. This results in the solution having a H closer to that of water.

If one adds a very small amount of strong base to a buffered solution, the pH will remain constant because a buffer solution acts to withstand any change to its pH on the addition of small quantities of either an acid or a base.

A buffer solution cannot be made with a strong acid because thy undergo complete dissociation. Therefore, any small addition of base or acid will result in very large changes in the pH of the solution. A buffer solution is made with a weak acid and its conjugate base or a weak base and its conjugate acid.

6 0
3 years ago
Can someone please help me on this
Fofino [41]

Answer:

it's none because the graph is decreasing like a milf so that why

7 0
3 years ago
Has the spring been stretched beyond its elastic limit?Explain your answer​
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4 0
3 years ago
Strotium−90, a radioactive isotope, is a major product of an atomic bomb explosion. It has a half-life of 28.1 yr. (a) Calculate
Bess [88]

Answer :

(a) The first-order rate constant for the nuclear decay is, 0.0247\text{ years}^{-1}

(b) The fraction of 90-Sr that remains after 10 half-lives is, \frac{a_o}{1024}

(c) The time passed in years is, 127.4 years.

Explanation :

<u>Part (a) :</u>

Half-life = 28.1 years

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{28.1\text{ years}}

k=0.0247\text{ years}^{-1}

Thus, the first-order rate constant for the nuclear decay is, 0.0247\text{ years}^{-1}

<u>Part (b) :</u>

Now we have to calculate the fraction of 90-Sr that remains after 10 half-lives.

Formula used :

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives

a_o = Initial amount of the reactant

n = number of half lives  = 10

Now put all the given values in the above formula, we get:

a=\frac{a_o}{2^{10}}

a=\frac{a_o}{1024}

Thus, the fraction of 90-Sr that remains after 10 half-lives is, \frac{a_o}{1024}

<u>Part (c) :</u>

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 0.0247\text{ years}^{-1}

t = time passed by the sample  = ?

a = let initial amount of the reactant  = 100

a - x = amount left after decay process = 100 - 95.7 = 4.3

Now put all the given values in above equation, we get

t=\frac{2.303}{0.0247}\log\frac{100}{4.3}

t=127.4\text{ years}

Therefore, the time passed in years is, 127.4 years.

6 0
3 years ago
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