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umka21 [38]
1 year ago
8

Why is the center of the buffer region of a weak acid–strong base titration significant?

Chemistry
1 answer:
Svetradugi [14.3K]1 year ago
7 0

When a weak acid is titrated against a strong base, the pH of the solution rises, levels off through the buffer zone, and then rises quickly to reach the equivalence point.

<h3>What is titration?</h3>

Titration is a typical research center technique for quantitative investigation synthetic  to decide the convergence of a recognized analyte. Titration is a crucial cycle in the creation of cheddar. For this situation, titration is significant while estimating the pH and acridity of the underlying milk being utilized to make the cheddar. Corrosive base titration is a quantitative investigation of convergence of an obscure corrosive or base arrangement. Titration, otherwise called titrimetry, is a typical lab strategy for quantitative substance investigation that is utilized to decide the convergence of a distinguished analyte. A reagent, called the titrant or titrator is ready as a standard arrangement. A known fixation and volume of titrant responds with an answer of analyte or titrand to decide focus. The volume of titrant responded is called titration volume.

Learn more about titration, visit

brainly.com/question/22304973

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33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b
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Explanation:

Mass of fructose = 33.56 g

Mass of water =  18.88  g

Total mass of the solution =  Mass of fructose + Mass of water = M

M = 33.56 g + 18.88  g =52.44 g

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Density =\frac{Mass}{Volume}

a) Density of the solution:

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b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol

Molar mass of water = 18.02 g/mol

Moles of water= n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol

Mole fraction of fructose in this solution:\chi_1

\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}

\chi_1=0.1510

Mole fraction of water = \chi_2=1-\chi_1=0.8490

c) Average molar mass of of the solution:

=\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol

=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

\frac{\text{Average molar mass}}{\text{Density of the mass}}

=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol

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