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Chapter 14: Problem 50

Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for each solution at \(25^{\circ} \mathrm{C}\). Identify each solution as neutral, acidic, or basic. a. \(\mathrm{pH}=7.40\) (the normal \(\mathrm{pH}\) of blood) b. \(\mathrm{pII}=15.3\) c. \(\mathrm{pH}=-1.0\) d. \(\mathrm{pH}=3.20\) e. \(\mathrm{pOH}=5.0\) f. \(\mathrm{pOH}=9.60\)

Short Answer

Expert verified
a. \(\left[\mathrm{H}^{+}\right] = 3.98 \times 10^{-8} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 2.51 \times 10^{-7} \mathrm{M}\), basic. b. \(\left[\mathrm{H}^{+}\right] = 5.01 \times 10^{-16} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 2 \times 10^2 \mathrm{M}\), basic. c. \(\left[\mathrm{H}^{+}\right] = 1 \times 10^1 \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 1 \times 10^{-15} \mathrm{M}\), acidic. d. \(\left[\mathrm{H}^{+}\right] = 6.31 \times 10^{-4} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 1.58 \times 10^{-11} \mathrm{M}\), acidic. e. \(\left[\mathrm{H}^{+}\right] = 1 \times 10^{-9} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 1 \times 10^{-5} \mathrm{M}\), basic. f. \(\left[\mathrm{H}^{+}\right] = 2.51 \times 10^{-5} \mathrm{M}\), \(\left[\mathrm{OH}^{-}\right] = 3.98 \times 10^{-10} \mathrm{M}\), acidic.

Step by step solution

01

Calculate \(\left[\mathrm{H}^{+}\right]\) for a given pH value

To calculate the concentration of hydrogen ions (\(\left[\mathrm{H}^{+}\right]\)), we use the formula: \[\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}}\] So, \[\left[\mathrm{H}^{+}\right] = 10^{-7.40} = 3.98 \times 10^{-8} \mathrm{M}\]
02

Calculate pOH and \(\left[\mathrm{OH}^{-}\right]\)

We first calculate pOH using the relationship \(pK_w = pH + pOH\). Therefore, \(pOH = pK_w - pH = 14 - 7.40 = 6.60\). Now, to calculate the concentration of hydroxide ions (\(\left[\mathrm{OH}^{-}\right]\)), we use the formula: \[\left[\mathrm{OH}^{-}\right] = 10^{-\mathrm{pOH}}\] So, \[\left[\mathrm{OH}^{-}\right] = 10^{-6.60} = 2.51 \times 10^{-7} \mathrm{M}\]
03

Identify as neutral, acidic, or basic

Since the value of the given pH (7.40) is greater than 7, the solution is basic. b. \(\mathrm{pH}=15.3\) Repeat the steps from solution a. We calculate \(\left[\mathrm{H}^{+}\right]\), pOH, and \(\left[\mathrm{OH}^{-}\right]\). Since the value of the given pH (15.3) is greater than 7, the solution is basic. c. \(\mathrm{pH}=-1.0\) Repeat the steps from solution a. We calculate \(\left[\mathrm{H}^{+}\right]\), pOH and \(\left[\mathrm{OH}^{-}\right]\). Since the value of the given pH (-1.0) is less than 7, the solution is acidic. d. \(\mathrm{pH}=3.20\) Repeat the steps from solution a. We calculate \(\left[\mathrm{H}^{+}\right]\), pOH and \(\left[\mathrm{OH}^{-}\right]\). Since the value of the given pH (3.20) is less than 7, the solution is acidic. e. \(\mathrm{pOH}=5.0\)
04

Calculate pH and \(\left[\mathrm{H}^{+}\right]\)

We first calculate pH using the relationship \(pK_w = pH + pOH\). Therefore, \(pH = pK_w - pOH = 14 - 5.0 = 9.0\). Now, to calculate the concentration of hydrogen ions (\(\left[\mathrm{H}^{+}\right]\)), we use the formula: \[\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}}\] So, \[\left[\mathrm{H}^{+}\right] = 10^{-9.0} = 1 \times 10^{-9} \mathrm{M}\]
05

Calculate \(\left[\mathrm{OH}^{-}\right]\)

To calculate the concentration of hydroxide ions (\(\left[\mathrm{OH}^{-}\right]\)), we use the formula: \[\left[\mathrm{OH}^{-}\right] = 10^{-\mathrm{pOH}}\] So, \[\left[\mathrm{OH}^{-}\right] = 10^{-5.0} = 1 \times 10^{-5} \mathrm{M}\]
06

Identify as neutral, acidic, or basic

Since the value of the calculated pH (9.0) is greater than 7, the solution is basic. f. \(\mathrm{pOH}=9.60\) Repeat the steps from solution e. We calculate pH, \(\left[\mathrm{H}^{+}\right]\), and \(\left[\mathrm{OH}^{-}\right]\). Since the value of the calculated pH is less than 7, the solution is acidic.

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Key Concepts

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

Acidic Solutions
Acidic solutions are those that possess a higher concentration of hydrogen ions (\([\mathrm{H}^{+}]\)) than hydroxide ions. In the pH scale, which ranges from 0 to 14, solutions with a pH less than 7 are considered acidic. This is because the formula \( [\mathrm{H}^{+}] = 10^{-\mathrm{pH}} \) results in higher hydrogen ion concentrations when pH values are lower.

Some common examples of acidic solutions include vinegar, lemon juice, and hydrochloric acid. They tend to have a sour taste and can be corrosive.
  • The lower the pH, the higher the acidity.
  • For example, a solution with a pH of 3 has 1000 times more \([\mathrm{H}^{+}]\) than pure water.
Basic Solutions
Basic solutions, also referred to as alkaline solutions, have a higher concentration of hydroxide ions (\([\mathrm{OH}^{-}]\)) compared to hydrogen ions. On the pH scale, solutions with a pH greater than 7 are classified as basic. The calculation \([\mathrm{OH}^{-}] = 10^{-\mathrm{pOH}} \) helps determine the concentration of hydroxide ions.

Common examples include soap, baking soda solutions, and household ammonia. These solutions usually feel slippery to the touch and can neutralize acids.
  • Higher pH numbers indicate increased basicity.
  • A solution with a pH of 9 has a lower hydrogen ion concentration compared to a neutral solution.
Neutral Solutions
Neutral solutions have equal concentrations of hydrogen and hydroxide ions. Pure water is a classic example, existing at pH 7. This balance is described by the equation:\([\mathrm{H}^{+}] = [\mathrm{OH}^{-}]\).

Though the pH of water remains 7 under normal circumstances, when mixed with other substances, this balance might shift, causing the pH to change. Regardless, the fundamental definition of neutrality relies on having identical concentrations of \([\mathrm{H}^{+}]\) and \([\mathrm{OH}^{-}]\).
  • This equal concentration is consistent only in distilled water or when there are no additional acid or base agents present.
  • Neutral solutions are neither acidic nor basic.
Hydrogen Ion Concentration
Hydrogen ion concentration, denoted as \([\mathrm{H}^{+}]\), is crucial in determining the acidity of a solution. The concentration is directly derived from the pH value using \([\mathrm{H}^{+}] = 10^{-\mathrm{pH}}\).

When \([\mathrm{H}^{+}]\) is higher, acidic characteristics of the solution become more pronounced. For instance, a pH of 1 results in a hydrogen ion concentration of \(1 \times 10^{-1}\) M, indicating a strong acid.
  • The relationship is exponential, meaning each unit change in pH represents a tenfold change in \([\mathrm{H}^{+}]\)
  • This makes small pH differences significant when comparing acidity levels.
Hydroxide Ion Concentration
The concentration of hydroxide ions, represented as \([\mathrm{OH}^{-}]\), indicates how basic a solution is. This value can be calculated using the formula \([\mathrm{OH}^{-}] = 10^{-\mathrm{pOH}}\).

For example, a pOH of 3 yields a hydroxide ion concentration of \(1 \times 10^{-3}\) M, showing a strong base. This concentration is inversely related to hydrogen ion concentration because \(pH + pOH = 14\).
  • Higher \([\mathrm{OH}^{-}]\) means a more basic or alkaline solution.
  • Basic solutions have lower \([\mathrm{H}^{+}]\) values.

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Most popular questions from this chapter

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a \(3.0 \times 10^{-7}-M\) solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

Calculate \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right]\), and the \(\mathrm{pH}\) of \(0.40 M\) solutions of each of the following amines (the \(K_{\mathrm{b}}\) values are found in Table \(14.3\) ). a. aniline b. methylamine

Write out the stepwise \(K_{\mathrm{a}}\) reactions for citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\), a triprotic acid.

Calculate \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right]\), and the \(\mathrm{pH}\) of \(0.20 \mathrm{M}\) solutions of each of the following amines. a. triethylamine \(\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{~N}, K_{\mathrm{b}}=4.0 \times 10^{-4}\right]\) b. hydroxylamine \(\left(\mathrm{HONH}_{2}, K_{\mathrm{b}}=1.1 \times 10^{-8}\right)\)

Calculate the \(\mathrm{pH}\) of a \(5.0 \times 10^{-3}-M\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\).
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