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2 years ago

3

The ratio of the concentration of a(n) over describes the proportions of a weak acid necessary to satisfy the

henderson-hasselbalch equation.

2 answers:

Vladimir79 [104]2 years ago

6 0

The pH of a buffer is calculated using Hendersen Hassalbalch's equation

pH = pKa + log [salt] / [acid]

thus ratio of concentration of conjugate base or species which can accept proton over a proton donor or conjugate acid describes the proportion of weak acid.

The solution of weak acid with its salt of strong base results into formation of a buffer, solution which resists change in pH on addition of small amount of strong acid or strong base

svp [43]2 years ago

3 0

Answer:

The ratio of the concentration of a <em><u>conjugate base</u></em> over <em><u>weak acid </u></em>describes the proportions of a weak acid necessary to satisfy the Henderson-Hasselbalch equation.

Explanation:

The Henderson Hesselbach equation is given as:

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

pH = potential of H

$pK_a=-\log [K_a]$

$K_a$ =Dissociation constant

[Salt] = Concentration of salt

[Acid] = Concentration of acid

Or it can also be represented as:

WeaK acid $\rightleftharpoons H^+$ + Conjugate base

$pH=pK_a+\log \frac{[\text{conjugate base}]}{[\text{weak acid}]}$

[conjugate base] = Concentration of conjugate base formed from the dissociation of weak acid.

[weak acid] = Concentration of weak acid.

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Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces

Shalnov [3]

Answer: The concentration of excess $[OH^-]$ in solution is 0.017 M.

Explanation:

1. $Molarity=\frac{moles}{\text {Volume in L}}$

moles of $HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles$

1 mole of $HCl$ give = 1 mole of $H^+$

Thus 0.019 moles of $HCl$ give = 0.019 mole of $H^+$

2. moles of $Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles$

According to stoichiometry:

1 mole of $Ba(OH)_2$ gives = 2 moles of $OH^-$

Thus 0.012 moles of $Ba(OH)_2$ give = $2 \times 0.012=0.024$ moles of $OH^-$

$H^++OH^-\rightarrow H_2O$

As 1 mole of $H^+$ neutralize 1 mole of $OH^-$

0.019 mole of $H^+$ will neutralize 0.019 mole of $OH^-$

Thus (0.024-0.019)= 0.005 moles of $OH^-$ will be left.

$[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M$

Thus molarity of $[OH^-]$ in solution is 0.017 M.

4 0

2 years ago

Read 2 more answers

HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.

Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, $C_A$ = 6.0mol/dm³

volume of HA, $V_A$ = 25.0cm³, = 0.025dm³

Concentration of HB, $C_B$ = 3.0mol/dm³

volume of HB, $V_B$ = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

$C_iVi = C_fV_f$

where;

$C_i$ is initial concentration

$V_i$ is initial volume

$C_f$ is final concentration of the solution

$V_f$ is final volume of the solution

$C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3$

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0

2 years ago

When monomers combine to form a condensation polymer, another product is also formed. typically, this product is _____. methanol

shtirl [24]

Water is the answer!

4 0

2 years ago

Read 2 more answers

Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O → 12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0

2 years ago

Sati [7]

In a chemical reaction, the limiting reagent is the chemical being used up while the excess reactant is the chemical left after the reaction process.

Before calculating the limiting and excess reactant, it is important to balance the equation first by stoichiometry.

C25N3H30Cl + NaOH = C25N3H30OH + NaCl

Since the reaction is already balanced, we can now identify which is the limiting and excess reagent.

First, we need to determine the number of moles of each chemical in the equation. This is crucial for determining the limiting and excess reagent.

<span>Assuming that there is the same amount of solution X for each reactant</span>

1.0 M NaOH ( X ) = 1.0 moles NaOH

1.00 x 10-5 M C25N3H30Cl ( X ) = 1.00 x 10-5 moles C25N3H30Cl

<span>The result showed that the crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this chemical reaction is crystal violet and the excess reactant is NaOH.</span>

3 0

2 years ago