Problem 106
Question
(graph can't copy) Suppose that \(x\) liters of pure acid are added to 200 liters of a \(35 \%\) acid solution. a. Write a formula that gives the concentration, \(C,\) of the new mixture. (Hint: Sce Exercise 105 .) b. How many liters of pure acid should be added to produce a new mixture that is \(74 \%\) acid?
Step-by-Step Solution
Verified Answer
Approximately 300 liters of pure acid needs to be added to get 74% acid solution.
1Step 1: Formula for the Concentration
Given a 200 liters solution with a concentration of 35%, we have 200 * 0.35 = 70 liters of acid. Now if x liters of pure acid are added, the total volume of the solution becomes (200 + x) liters and the total volume of acid becomes (70 + x) liters. So the concentration of the new mixture, \(C\), is given by \[C = \frac{{70 + x}}{{200 + x}}\]
2Step 2: Solving for the Required Acid
We want the final concentration to be 74% or 0.74. Plugging this value to the formula from Step 1, we have \[0.74 = \frac{{70 + x}}{{200 + x}}\]. We can solve this equation for x.
3Step 3: Solve the Equation
Cross multiplying gives us \[0.74 (200 + x ) = 70 + x\]. Simplifying this equation we get, \[148 + 0.74x = 70 + x\]. Subtracting 0.74x from both sides we got, \[148 - 70 = x - 0.74x\], so \[78 = 0.26x\]. Solving for x, \[x = \frac{78}{0.26} ≈ 300\], so approximately 300 liters of pure acid needs to be added.
Key Concepts
Pure Acid AdditionPercent ConcentrationSolution VolumeAlgebraic Equations
Pure Acid Addition
When trying to increase the concentration of an acid solution, one effective method is to add pure acid directly. This addition increases both the total volume and the concentration of acid in the solution. Let's break down what happens:
- Suppose you start with a solution, like in our exercise, that originally contains a certain percentage of acid mixed with other components, usually water.
- By adding pure acid, you are directly increasing the quantity of acid present in the solution without diluting it with non-acid components.
- This impacts the overall concentration because pure acid has 100% acid content, directly pushing up the average concentration of acid in the mix.
Percent Concentration
Percent concentration is a way of expressing how much of a particular solute is present in a solution. In the context of our problem, it refers to the amount of acid in the total solution. Here's how you can think about percent concentration:
- It is calculated by dividing the volume of solute (in this case, acid) by the total volume of the solution, then multiplying by 100 to get a percentage.
- Therefore, if a solution has a 35% acid concentration, it means 35% of the total solution volume is composed of acid.
- This numerical representation helps you quickly understand the strength or potency of the solution.
Solution Volume
Solution volume is an important factor when calculating concentration, as it represents the total space occupied by the mixture. It’s easy to see how it works with our exercise:
- The initial solution volume is 200 liters, containing some amount of pure acid and some amount of other components.
- When you add pure acid, the total volume increases by the amount of added acid. Thus, the new volume becomes the sum of the initial volume and added pure acid.
- Understanding the relationship between initial and final volume helps clarify how the concentration of a solution changes when pure substances are added or removed.
Algebraic Equations
Algebraic equations are powerful tools for solving problems involving concentrations and mixtures. Let's dive into how they are used in our exercise:
- We start with a given equation for concentration: \[C = \frac{{70 + x}}{{200 + x}}\]. Here, 70 is the amount of acid in liters, and \(x\) is the liters of pure acid added.
- We solve for \(x\) by setting the concentration, \(C\), equal to the desired percentage expressed as a decimal, like 0.74 for 74%.
- Cross-multiplication helps eliminate the fraction, simplifying it to a solvable linear equation in terms of \(x\).
- The solution gives the exact amount of pure acid needed to reach the desired concentration.
Other exercises in this chapter
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