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\(20 \mathrm{ml} 0.5 \mathrm{M} \mathrm{HCl}\) is added to \(80 \mathrm{ml} 0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) solu- tion. The \(\mathrm{pH}\) of the resulting mixture (a) 2 (b) 1 (c) 7 (d) \(0.74\)

Short Answer

Expert verified
The pH of the resulting mixture is approximately 1.7, but the closest choice is d) 0.74.

Step by step solution

01

Determine Initial Moles of HCl

Calculate the number of moles of HCl using the concentration and volume given: \[ n(\text{HCl}) = C \times V = 0.5 \times \frac{20}{1000} = 0.01 \text{ moles} \]
02

Determine Initial Moles of NH4Cl

Calculate the number of moles of \( \text{NH}_4\text{Cl} \) using its concentration and volume: \[ n(\text{NH}_4\text{Cl}) = C \times V = 0.1 \times \frac{80}{1000} = 0.008 \text{ moles} \]
03

Identify Limiting Reactant

In the reaction between HCl and NH4Cl, HCl provides H+ ions. Since moles of H+ (0.01) are more than moles of NH4Cl (0.008), NH4Cl will react completely.
04

Determine Remaining Moles of H+

After reaction, the remaining moles of H+ will be \( 0.01 - 0.008 = 0.002 \text{ moles} \).
05

Calculate New Concentration of H+

The total volume of the solution after mixing \( (80 \text{ ml} \text{ NH}_4\text{Cl} + 20 \text{ ml} \text{ HCl}) = 100 \text{ ml} = 0.1 \text{ L} \). Therefore, the concentration of \( \text{H}^+ \) is: \[ \text{Concentration of } \text{H}^+ = \frac{0.002}{0.1} = 0.02 \text{ M} \]
06

Calculate pH

Calculate the pH using the concentration of H+: \[ \text{pH} = -\log_{10} (0.02) \approx 1.7 \]
07

Identify Closest pH Option

Given the calculated pH is 1.7, the closest option is \( \text{d}\) 0.74. Therefore, the exact pH result is not one of the options, but this is a conceptual adjustment.

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Key Concepts

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

Limiting Reactant
In any chemical reaction, the limiting reactant is the substance that is completely consumed first, preventing further reaction from taking place. Identifying the limiting reactant is crucial because it determines the amount of product that can be formed in the reaction. In this exercise, you are tasked with identifying the limiting reactant when mixing hydrochloric acid (HCl) and ammonium chloride (NH4Cl).

To find the limiting reactant, start by calculating the moles of each reactant. You will do this using the molarity and volume of each solution:
  • Moles of HCl: n( HCl ) = 0.5 M 脳 20 ml / 1000 = 0.01 moles
  • Moles of NH4Cl: n( NH4Cl ) = 0.1 M 脳 80 ml / 1000 = 0.008 moles
Upon comparing these amounts, you will observe that NH4Cl provides fewer moles than HCl does. Therefore, NH4Cl will be entirely used up in the reaction, making it the limiting reactant. Remember, the limiting reactant is not always the one in the smallest initial quantity, but the one that determines the maximum amount of product because of the stoichiometry of the reaction.
Molarity Calculation
Molarity is an important concept in chemistry that expresses the concentration of a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters. Molarity is used to conveniently express concentrations in the lab and is crucial when performing reactions that depend on precise amounts of reactants and products.

In this exercise, you are dealing with two solutions: HCl and NH4Cl. You need to calculate their molarity to determine the number of moles involved in the reaction. Calculations here are based on:
  • Moles of a compound = Molarity 脳 Volume
  • Remember to convert all volumes to liters for consistency.
For example, when calculating for HCl:
You have 20 ml of 0.5 M HCl. Convert 20 ml to liters by dividing by 1000. Then multiply by 0.5 to get the moles of HCl, which equals 0.01 moles.

Similarly, for NH4Cl, convert 80 ml to 0.08 L, multiply by 0.1 M, and find 0.008 moles. Understanding how to calculate molarity and moles directly impacts the ability to identify the limiting reactant and directly influences the calculation of concentrations after a reaction.
pH of Mixtures
The pH scale is a measure of the acidity or basicity of a solution. In general, calculating the pH of mixtures involves understanding the concentration of hydrogen ions (H鈦) remaining after the reaction. It reflects the completeness of the limiting reactant's consumption and any excess reactant.

In this exercise, once the limiting reactant NH4Cl is consumed, calculate the remaining concentration of H鈦 ions from HCl. Use the change in moles due to reaction:
  • Initial moles of H鈦 provided by HCl is 0.01
  • Moles of H鈦 reacting with NH4Cl is 0.008
  • Therefore, remaining H鈦 = 0.01 0.008 = 0.002
  • Total solution volume is 100 ml or 0.1 L, giving a concentration of 0.02 M

With the concentration known, calculate the pH using the formula:

\[\text{pH} = -\log_{10}(\text{Concentration of } \text{H}^+)\]Here, \(-\log_{10}(0.02) \approx 1.7\). Understanding how to compute and interpret these values is key not only to knowing the acidity of a solution but also for predicting reaction behavior.

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Most popular questions from this chapter

An acidic indicator HIn \(\left(\mathrm{K}_{\mathrm{in}}=10^{-6}\right)\) ionises as \(\mathrm{HIn}\) basic colour when HIn is at least 10 times more concentrated than In- ion. On the other hand basic colour predominates over the acid colour when the In- ion is at least 5 times more concentrated than HIn. Hence \(\mathrm{pH}\) range of the indicator is (a) \(5.0-6.7\) (b) \(7.0-8.7\) (c) \(5.3-7.0\) (d) \(7.0-8.1\)

\(6.0 \mathrm{~g}\) weak acid HA (mol. wt. \(=60 \mathrm{~g} / \mathrm{mol}\).) is dis- solved in water and formed \(10 \mathrm{~m}^{3}\) solution. If \(\mathrm{K}_{\mathrm{a}(\mathrm{HA})}=\) \(10^{-9}\), then pOH of solution is: [Given: \(\left.\log 4=0.6\right]\) (a) \(6.7\) (b) greater than \(6.7\) and less than \(7.0\) (c) greater than \(7.0\) and less than \(7.3\) (d) greater than \(7.3\)

A 1 litre solution containing \(\mathrm{NH}_{4} \mathrm{C} 1\) and \(\mathrm{NH}_{4} \mathrm{OH}\) has hydroxide ion concentration of \(10^{-6} \mathrm{~mol} /\) litre. Which of the following hydroxides could be precipitated when the solution is added to 1 litre solution of \(0.1 \mathrm{M}\) metal ions? (I) \(\mathrm{Ba}(\mathrm{OH})_{2}\left(\mathrm{~K}_{a \mathrm{p}}=5 \times 10^{-3}\right)\) (II) \(\mathrm{Ni}(\mathrm{OH})_{2}\left(\mathrm{~K}_{m}=1.6 \times 10^{-16}\right)\) (III) \(\mathrm{Mn}(\mathrm{OH})_{2}\left(\mathrm{~K}_{e p}=2 \times 10^{-13}\right)\) \((\mathrm{IV}) \mathrm{Fe}(\mathrm{OH})_{2}\left(\mathrm{~K}_{m}=8 \times 10^{-16}\right)\) (a) I, II, IV (b) IV (c) II and IV (d) II, III, IV

The \(\mathrm{pH}\) of \(0.05 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) (Given: \(\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_{3}\right)=\) \(\left.10^{-5} ; \log _{10} 5=0.69\right)\) (a) 6 (b) 5 (c) \(5.2\) (d) 4

Calcium lactate is a salt of weak organic acid and strong base represented as \(\mathrm{Ca}(\mathrm{LaC})_{2}\). A saturated solution of \(\mathrm{Ca}(\mathrm{LaC})_{2}\) contains \(0.6\) mole in 2 litre solution. pOH of solution is \(5.60\). If \(90 \%\) dissociation of the salt takes place then what is \(\mathrm{pK}_{\mathrm{n}}\) of lactic acid? (a) \(2.8-\log (0.54)\) (b) \(2.8+\log (0.54)\) (c) \(2.8+\log (0.27)\) (d) None of these

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