Chemistry, 12.07.2019 11:30 brisamauro27

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? 14) a.2.85 b.0.85 c.2.96 d.3.81 e.1.49

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Answer from: shelbybibb99

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 ? link

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

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: SmokeyRN

b. 0.85

Step 1. Write the chemical equation

HCl + NaOH → NaCl + H2O

Step 2. Calculate the initial moles of HCl

Moles of HCl = 50.0 mL HCl × (0.50 mmol HCl/1 mL HCl) = 25.0 mmol HCl

Step 3. Calculate the moles of NaOH added

Moles of NaOH added = 28.0 mL NaOH × (0.50 mmol NaOH/(1 mL NaOH)

= 14.0 mmol NaOH

Step 4. Calculate the moles of HCl reacted

Moles of HCl reacted = 14.0 mmol NaOH × (1 mmol HCl/1 mmol NaOH)

= 14.0 mmol HCl

Step 5. Calculate the moles of HCl remaining

Moles of HCl remaining = (25.0 – 14.0) mmol HCl = 11.0 mmol HCl

Step 6. Calculate the total volume of the solution

Volume = (50.0 + 28.0) mL = 78.0 mL

Step 7. Calculate [H^(+)]

[H^(+)] = [HCl] = moles/litres = 11.0 mmol/78.0 mL = 0.141 mol/L

Step 8. Calculate the pH

pH = -log[H^(+)] = -log0.141 = 0.85

Answer from: Quest

Yes you are correct use the scientific method

Answer from: Quest

the dissociation of nickel hydroxide is follows:

$ni(oh)_2 > ni^2^+ + 2 oh^-$

$ksp for no(oh)2 = 5.5 * 10^-^1^6$

$5.5 * 10^-^1^6 = s * (2 s)^2 = 4 s^3$

$s^3 = 137.5 * 10^-^1^8$

$s = 5.14 * 10^-^6 = [oh^-]$

the concentration of hydroxide ion is

$5.14 * 10^-^6 = [oh^-]$

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

$k = [ ni^2^+] * [oh^-]^2$

$6.1 * 10^-^1^6 = [ ni^2] * [5.14 * 10^-^6]^2$

$[ ni^2^+] = 0.23 * 10^-^4 mol /l$

thus, concentration of ni is 0.23 * 10^-^4 mol /l

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