Respuesta :
the balanced reaction between base and acid is;
NaOH + HCl ---> NaCl + H₂O
NaOH is a strong base and HCl is a strong acid therefore complete dissociation. Stoichiometry of acid to base is 1:1
the number of moles of base added - 0.5 M /1000 mL/L x 28.0 mL = 0.014 mol
the number of acid moles present - 0.5 M /1000 mL/L x 50.0 mL = 0.025 mol
acid reacts with base 1:1 ratio
therefore excess amount of acid present - 0.025 - 0.014 = 0.011 mol
total volume = 50.0 + 28.0 = 78.0 mL
[H⁺] = 0.011 mol/0.078 L = 0.14 M
pH = -log [H⁺]
pH = -log(0.14)
pH = 0.85
NaOH + HCl ---> NaCl + H₂O
NaOH is a strong base and HCl is a strong acid therefore complete dissociation. Stoichiometry of acid to base is 1:1
the number of moles of base added - 0.5 M /1000 mL/L x 28.0 mL = 0.014 mol
the number of acid moles present - 0.5 M /1000 mL/L x 50.0 mL = 0.025 mol
acid reacts with base 1:1 ratio
therefore excess amount of acid present - 0.025 - 0.014 = 0.011 mol
total volume = 50.0 + 28.0 = 78.0 mL
[H⁺] = 0.011 mol/0.078 L = 0.14 M
pH = -log [H⁺]
pH = -log(0.14)
pH = 0.85
The pH of the solution after addition of 28 mL of NaOH is added to HCl is [tex]\boxed{{\text{0}}{\text{.85}}}[/tex].
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:
[tex]{\text{Molarity of solution}}=\dfrac{{{\text{Moles}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of solution}}}}[/tex]
......(1)
Rearrange equation (1) to calculate the moles of solute.
[tex]{\text{Moles}}\;{\text{of}}\;{\text{solute}}=\left( {{\text{Molarity of solution}}} \right)\left( {{\text{Volume of solution}}} \right)[/tex] ......(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.
[tex]\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}[/tex]
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.
[tex]\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}[/tex]
The reaction between HCl and NaOH occurs as follows:
[tex]{\text{NaOH}} + {\text{HCl}} \to {\text{NaCl}} + {{\text{H}}_2}{\text{O}}[/tex]
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:
[tex]\begin{aligned}{\text{Amount of HCl remaining}}&= 0.02{\text{5 mol}} - 0.01{\text{4 mol}}\\&={\text{0}}{\text{.011 mol}}\\\end{aligned}[/tex]
The volume after the addition of NaOH can be calculated as follows:
[tex]\begin{aligned}{\text{Volume of solution}}&= {\text{50 mL}} + {\text{28 mL}}\\&= {\text{78 mL}}\\\end{aligned}[/tex]
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.
[tex]\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}[/tex]
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:
[tex]{\text{pH}}=- {\text{log}}\left[{{{\text{H}}^ + }}\right][/tex] ......(3)
Here,
[tex]\left[{{{\text{H}}^ + }}\right][/tex] is hydrogen ion concentration.
HCl is a strong acid so it dissociates completely. So the concentration of [tex]{{\text{H}}^ + }[/tex] also becomes 0.141 M.
Substitute 0.141 M for [tex]\left[{{{\text{H}}^ + }}\right][/tex] in equation (3).
[tex]\begin{aligned}{\text{pH}}&= - {\text{log}}\left( {0.141} \right)\\&= 0.85\\\end{aligned}[/tex]
So the pH of the solution is 0.85.
Learn more:
1. Which indicator is best for titration between HI and ? https://brainly.com/question/9236274
2. Why is bromophenol blue used as an indicator for antacid titration? https://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.
