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Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of the following solutions. (a) \(0.0045 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) (b) \(0.080 \mathrm{MHI}\) (c) \(0.030 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) (d) \(12.3 \mathrm{M} \mathrm{HNO}_{3}\)

Short Answer

Expert verified
(a) pH 11.95, pOH 2.05; (b) pH 1.10, pOH 12.90; (c) pH 12.78, pOH 1.22; (d) pH -1.09, pOH 15.09.

Step by step solution

01

Determine OH- and H+ Concentrations for Ba(OH)2

Ba(OH)2 dissociates completely in water to give Ba虏鈦 and 2OH鈦 ions. So, the concentration of OH鈦 ions is twice that of the Ba(OH)2 concentration. Given:- \(\mathrm{[Ba(OH)_2] = 0.0045\ M}\)- \(\mathrm{[OH^-] = 2 \times 0.0045\ M = 0.009\ M}\)
02

Calculate pOH and pH for Ba(OH)2

The pOH of a solution can be calculated using the concentration of OH鈦:\(\mathrm{pOH = -\log[OH^-]}\)\(\mathrm{pOH = -\log(0.009) \approx 2.05}\)Next, use the relationship between pH and pOH: \(\mathrm{pH + pOH = 14}\)\(\mathrm{pH = 14 - 2.05 = 11.95}\)
03

Determine H+ Concentration for HI

HI is a strong acid and dissociates completely in water. Therefore, the concentration of H鈦 ions is equal to the HI concentration.Given:- \(\mathrm{[HI] = 0.080\ M}\)- \(\mathrm{[H^+] = 0.080\ M}\)
04

Calculate pH and pOH for HI

The pH of a solution can be calculated using the concentration of H鈦:\(\mathrm{pH = -\log[H^+]}\)\(\mathrm{pH = -\log(0.080) \approx 1.10}\)Then, find the pOH:\(\mathrm{pH + pOH = 14}\)\(\mathrm{pOH = 14 - 1.10 = 12.90}\)
05

Determine OH- Concentration for Sr(OH)2

Sr(OH)2 dissociates completely in water to give Sr虏鈦 and 2OH鈦 ions. So, the concentration of OH鈦 ions is twice that of the Sr(OH)2 concentration.Given:- \(\mathrm{[Sr(OH)_2] = 0.030\ M}\)- \(\mathrm{[OH^-] = 2 \times 0.030\ M = 0.060\ M}\)
06

Calculate pOH and pH for Sr(OH)2

The pOH can be computed using the OH鈦 concentration:\(\mathrm{pOH = -\log[OH^-]}\)\(\mathrm{pOH = -\log(0.060) \approx 1.22}\)Now, calculate the pH:\(\mathrm{pH = 14 - 1.22 = 12.78}\)
07

Determine H+ Concentration for HNO3

HNO3 is a strong acid and dissociates completely in water. Therefore, the concentration of H鈦 ions is equal to the HNO3 concentration.Given:- \(\mathrm{[HNO_3] = 12.3\ M}\)- \(\mathrm{[H^+] = 12.3\ M}\)
08

Calculate pH and pOH for HNO3

Find the pH using the concentration of H鈦:\(\mathrm{pH = -\log[H^+]}\)\(\mathrm{pH = -\log(12.3) \approx -1.09}\) (not common, but possible in very concentrated solutions)Then compute pOH:\(\mathrm{pH + pOH = 14}\)\(\mathrm{pOH = 14 - (-1.09) = 15.09}\)

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Key Concepts

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

Acid-Base Chemistry
In acid-base chemistry, it's essential to understand how acids and bases interact in a solution. This field involves studying substances that either donate protons (H鈦) or accept them. Acids like \( ext{HI}\) and \( ext{HNO}_3\) donate H鈦 ions in a solution, earning them the title of proton donors. Bases, like \( ext{Ba(OH)}_2\) and \( ext{Sr(OH)}_2\), typically accept protons, usually by providing hydroxide ions (OH鈦) that combine with H鈦 to form water.
The concept of pH is central in acid-base chemistry. It's a measure of the acidity or basicity of a solution. For acids, a low pH indicates a high concentration of H鈦 ions, while a high pH in bases indicates low H鈦 ion presence due to higher OH鈦 ion concentrations. The pH scale typically ranges from 0 to 14, with 7 as neutral. Below 7 is acidic, and above 7 is basic.
Solution Concentration
Concentration refers to the amount of a substance in a given volume of solution. In the context of acid-base reactions, it's crucial to know the concentration to calculate both pH and pOH. Having the molarity ( ext{M}) or concentration helps predict how much H鈦 or OH鈦 will be available in a solution. For instance:
  • \(0.080 \ ext{ M} \ ext{ HI}\) means 0.080 moles of HI in one liter of solution.
  • \(0.0045 \ ext{ M} \ ext{ Ba(OH)}_2\) suggests each liter of solution contains 0.0045 moles of Ba(OH)鈧.
Calculating the concentration of OH鈦 or H鈦 depends upon the degree of dissociation of compounds in water. For strong acids like HI and HNO鈧, the H鈦 concentration in water equals the initial concentration of the acid due to complete dissociation. Similarly, for strong bases like Ba(OH)鈧 and Sr(OH)鈧, the concentration of OH鈦 ions is often doubled due to dissociation.
Strong Acids and Bases
Strong acids and bases are vital components in the study of acid-base chemistry. These compounds dissociate completely in water, making it easy to calculate the concentration of ions they release.
Strong acids, such as HI and HNO鈧, readily release hydrogen ions (H鈦) when in aqueous solutions. For instance, a solution of 12.3 M HNO鈧 means the solution has 12.3 M H鈦 resulting from full dissociation. In terms of pH, such strong acids often result in very low pH values, sometimes below 0 for concentrated solutions.
Strong bases like \( ext{Ba(OH)}_2\) and \( ext{Sr(OH)}_2\) produce hydroxide ions (OH鈦) in water. For these bases, the concentration of OH鈦 is calculated by multiplying the initial concentration by the number of moles of OH鈦 ions each molecule releases. Thus, \(0.0045 \ ext{ M} \ ext{ Ba(OH)}_2\) will give 0.009 M OH鈦 ions in solution. Understanding these calculations is crucial for determining the pOH and linking it back to pH with \(\text{pH} + \text{pOH} = 14\).

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