/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 The pH of a buffer solution is \... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 10

The pH of a buffer solution is \(6.5\). The \(\mathrm{pK}_{\mathrm{a}}\) of the buffer is \(7.5\). If the total concentration of the buffer \(\left(\mathrm{HA}+\mathrm{A}^{-}\right)\) is 10 \(\mathrm{mmol} / \mathrm{L}\), what is the concentration of the conjugate base? a. \(0.0909 \mathrm{mmol} / \mathrm{L}\) b. \(0.909 \mathrm{mmol} / \mathrm{L}\) c. \(9 \mathrm{mmol} / \mathrm{L}\) d. \(90.9 \mathrm{mmol} / \mathrm{L}\) e. \(4.45 \mathrm{mmol} / \mathrm{L}\)

Short Answer

Expert verified
b. \(0.909 \mathrm{mmol/L}\)

Step by step solution

01

- Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used to relate the pH, \text{{pKa}}, and the ratio of the concentrations of the conjugate base \text{{(A鈦�)}} and the weak acid \text{{(HA)}}. The equation is given by: \[ \text{{pH}} = \text{{pKa}} + \text{{log}}\frac{[\text{{A}}^-]}{[\text{{HA}}]} \]
02

- Substitute the given values into the equation

Substitute \text{{pH}} = 6.5 and \text{{pKa}} = 7.5 into the Henderson-Hasselbalch equation: \[ 6.5 = 7.5 + \text{{log}}\frac{[\text{{A}}^-]}{[\text{{HA}}]} \]
03

- Solve for the ratio \text{{[A鈦籡/[HA]}}

Isolate the logarithmic term by subtracting 7.5 from both sides of the equation: \[ 6.5 - 7.5 = \text{{log}}\frac{[\text{{A}}^-]}{[\text{{HA}}]} \] This simplifies to: \[ -1 = \text{{log}}\frac{[\text{{A}}^-]}{[\text{{HA}}]} \]
04

- Exponentiate both sides to solve for the ratio

Exponentiate both sides using base 10 to remove the logarithm: \[ 10^{-1} = \frac{[\text{{A}}^-]}{[\text{{HA}}]} \] This simplifies to: \[ 0.1 = \frac{[\text{{A}}^-]}{[\text{{HA}}]} \]
05

- Express the concentrations in terms of a single variable

Let the concentration of \text{{A鈦粆} be \text{{x}}. Then the concentration of \text{{HA}} will be \text{{10 - x}} (because the total concentration is 10 \text{{mmol/L}}). Substitute these into the ratio: \[ 0.1 = \frac{x}{10 - x} \]
06

- Solve for x

Multiply both sides of the equation by \text{{10 - x}} to isolate \text{{x}}: \[ 0.1(10 - x) = x \] This simplifies to: \[ 1 - 0.1x = x \] Rearrange to get all terms involving \text{{x}} on one side: \[ 1 = 1.1x \] Solve for \text{{x}}: \[ x = \frac{1}{1.1} = 0.909 \text{{mmol/L}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

buffer solution pH calculation
To understand the pH of a buffer solution, we use the Henderson-Hasselbalch equation. This important equation helps us relate the pH, the acid dissociation constant (\text{{pKa}}), and the ratio of the concentrations of the buffer's conjugate base (\text{{A鈦粆}) and weak acid (\text{{HA}}). The equation for this is: \[ \text{{pH}} = \text{{pKa}} + \text{{log}}\frac{[\text{{A}}^-]}{[\text{{HA}}]} \] Using this equation, we can calculate the pH if we know the concentrations of \text{{A鈦粆} and \text{{HA}}, or vice versa. In the given exercise, we start by substituting the known values into the equation. If \text{{pH}} = 6.5 and \text{{pKa}} = 7.5, our goal is to find the concentration of the conjugate base or the weak acid.
acid-base equilibrium
Acid-base equilibrium is a fundamentally important concept in buffer solutions. It involves the balance between the concentrations of acids and bases in solution. For our buffer solution, we solve for the equilibrium by transforming the Henderson-Hasselbalch equation. \[ 6.5 = 7.5 + \text{{log}}\frac{[\text{{A}}^-]}{[\text{{HA}}]} \] By isolating and solving the logarithmic term, we find \[ -1 = \text{{log}}\frac{[\text{{A}}^-]}{[\text{{HA}}]} \] Applying exponentiation, we simplify the expression: \[ 10^{-1} = 0.1 = \frac{[\text{{A}}^-]}{[\text{{HA}}]} \] This ratio tells us how much conjugate base and acid are present at equilibrium. Understanding this balance is essential for predicting pH behavior in buffer solutions.
clinical chemistry calculations
In clinical chemistry, buffer solutions play a critical role in maintaining the pH stability of various bodily fluids. The accurate measurement and calculation of pH in buffers are essential for diagnosing and treating medical conditions. By using the ratio of the concentrations from our previous step \[ 0.1 = \frac{x}{10 - x} \] and solving for \text{{x}}, we find the concentration of the conjugate base. This involves rearranging the equation: \[ 0.1(10 - x) = x \] \[ 1 - 0.1x = x \] Solving for \text{{x}}, we get: \[ x = \frac{1}{1.1} = 0.909 \text{{mmol/L}} \] Thus, the concentration of our conjugate base \text{{A鈦粆} is 0.909 mmol/L. These calculations are necessary for ensuring buffer solutions in clinical settings are effective in maintaining pH consistency.

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