CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
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Chapter 19, Problem 19.129P

(a)

Interpretation Introduction

Interpretation:

Concentration of each species that is present like [HClO], [ClO], [H3O+], [OH], [Na+] has to be determined. 

Concept Introduction:

Dissociation constant of acid:

An acid dissociation constant, Ka quantitative measure of the strength of an acid in solution.  It is the equilibrium constant for a chemical reaction known as dissociation in the context of acid–base reactions. 

The chemical equation for acid dissociation can be written symbolically as:

HAA+H+

The acid dissociation constant can be presented as:

Ka=[A][H+][HA]

Concentration:

Concentration=No.ofmolesVolumeofsolution

(a)

Expert Solution
Check Mark

Answer to Problem 19.129P

[HClO], [ClO], [H3O+], [OH], [Na+] is calculated as 0.16M, 0.10M, 4.8×108M, 2.1×107M, and 0.10M respectively. 

Explanation of Solution

Given data:

Volumeofsolution=500mL=500×103L=0.500LMolofsolidNaOH=0.050MolofHClO=0.13KaofHCl=3.0×108

NaOH is a strong base, so it dissociates completely as follows:

NaOH(s)Na+(aq)+OH(aq)0.050mol0.050mol0.050mol

The OH ions from NaOH will react with HClO.

MolesOH(aq)+HClO(aq)H2O(l)+ClO(aq)_Initial0.0500.130Change0.0500.050+0.050_Equilibrium00.0800.050

[HClO]=0.080mol0.500L=0.16M.[ClO]=0.050mol0.500L=0.10M.

The balanced chemical equation of dissociation of HClO in water can be written as:

Concentration(M)HClO(aq)+H2O(l)H2O+(aq)+ClO(aq)_Initial0.160.10Changex+x+x_Equilibrium0.16xx0.10+x

As x is small, [HClO]0.16M and [ClO]0.10M.

Ka=[ClO][H3O+][HClO][H3O+]=[HClO]Ka[ClO]=(0.16)(3.0×108)0.10=4.8×108M.

Therefore, [H3O+] is 4.8×108M.

Kw=[H3O+][OH][OH]=Kw[H3O+]=1.0×10144.8×108=2.08333×107=2.1×107M.

Therefore, [OH] is 2.1×107M.

[Na+]=0.050mol0.500L=0.10M.

[Na+] is calculated as 0.10M.

(b)

Interpretation Introduction

Interpretation:

The pH of the solution has to be calculated. 

Concept Introduction:

pH=-log[H3O+]

(b)

Expert Solution
Check Mark

Answer to Problem 19.129P

The pH of the solution calculated as 7.32.

Explanation of Solution

[H3O+]=4.8×108MpH=log[H3O+]=log(4.8×108)=7.31875876=7.32.

Therefore, the pH of the solution calculated as 7.32.

(c)

Interpretation Introduction

Interpretation:

The pH after adding 0.0050molofHCl to the flask has to be determined. 

Concept Introduction:

pH=log[H3O+]

(c)

Expert Solution
Check Mark

Answer to Problem 19.129P

The pH after adding 0.0050molofHCl to the flask calculated as 7.25.

Explanation of Solution

Given data:

Volumeofsolution=500mL=500×103L=0.500LMolofsolidNaOH=0.050MolofHClO=0.13MolofHCl=0.0050KaofHCl=3.0×108

If 0.0050molofHCl is added then the ClO will react with the H+ to form HClO.

MolesH+(aq)+ClO(aq)HClO(aq)_Initial0.00500.0500.080Change0.00500.0050+0.0050_Equilibrium00.0450.085

[HClO]=0.085mol0.500L=0.17M.[ClO]=0.045mol0.500L=0.090M.

The balanced chemical equation of dissociation of HClO in water can be written as:

Concentration(M)HClO(aq)+H2O(l)H2O+(aq)+ClO(aq)_Initial0.170.090Changex+x+x_Equilibrium0.17xx0.090+x

As x is small, [HClO]0.17M and [ClO]0.090M.

Ka=[ClO][H3O+][HClO][H3O+]=[HClO]Ka[ClO]=(0.17)(3.0×108)0.090=5.6667×108M.

pH=log[H3O+]=log(5.6667×108)=7.246669779=7.25.

Therefore, the pH after adding 0.0050molofHCl to the flask calculated as 7.25.

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Chapter 19 Solutions

CHEMISTRY >CUSTOM<

Ch. 19.3 - Prob. 19.6AFPCh. 19.3 - Prob. 19.6BFPCh. 19.3 - Prob. 19.7AFPCh. 19.3 - Prob. 19.7BFPCh. 19.3 - Prob. 19.8AFPCh. 19.3 - Prob. 19.8BFPCh. 19.3 - Prob. 19.9AFPCh. 19.3 - Prob. 19.9BFPCh. 19.3 - Prob. 19.10AFPCh. 19.3 - Prob. 19.10BFPCh. 19.3 - Prob. 19.11AFPCh. 19.3 - Prob. 19.11BFPCh. 19.3 - Prob. 19.12AFPCh. 19.3 - Prob. 19.12BFPCh. 19.3 - An environmental technician collects a sample of...Ch. 19.3 - A lake that has a surface area of 10.0 acres (1...Ch. 19.4 - Cyanide ion is toxic because it forms stable...Ch. 19.4 - Prob. 19.13BFPCh. 19.4 - Prob. 19.14AFPCh. 19.4 - Calculate the solubility of PbCl2 in 0.75 M NaOH....Ch. 19 - Prob. 19.1PCh. 19 - Prob. 19.2PCh. 19 - Prob. 19.3PCh. 19 - Prob. 19.4PCh. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - Prob. 19.8PCh. 19 - Does the pH increase or decrease with each of the...Ch. 19 - The scenes below depict solutions of the same...Ch. 19 - The scenes below show three samples of a buffer...Ch. 19 - What are the [H3O+] and the pH of a propanoic...Ch. 19 - What are the [H3O+] and the pH of a benzoic...Ch. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Prob. 19.16PCh. 19 - Find the pH of a buffer that consists of 0.95 M...Ch. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Prob. 19.20PCh. 19 - Find the pH of a buffer that consists of 0.50 M...Ch. 19 - A buffer consists of 0.22 M KHCO3 and 0.37 M...Ch. 19 - A buffer consists of 0.50 M NaH2PO4 and 0.40 M...Ch. 19 - What is the component concentration ratio,...Ch. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - Prob. 19.28PCh. 19 - A buffer that contains 0.40 M of a base, B, and...Ch. 19 - A buffer that contains 0.110 M HY and 0.220 M Y−...Ch. 19 - A buffer that contains 1.05 M B and 0.750 M BH+...Ch. 19 - A buffer is prepared by mixing 204 mL of 0.452 M...Ch. 19 - A buffer is prepared by mixing 50.0 mL of 0.050 M...Ch. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Prob. 19.36PCh. 19 - Choose specific acid-base conjugate pairs to make...Ch. 19 - An industrial chemist studying bleaching and...Ch. 19 - Oxoanions of phosphorus are buffer components in...Ch. 19 - The scenes below depict the relative...Ch. 19 - Prob. 19.41PCh. 19 - What species are in the buffer region of a weak...Ch. 19 - Prob. 19.43PCh. 19 - Prob. 19.44PCh. 19 - Prob. 19.45PCh. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Prob. 19.53PCh. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Use figure 19.9 to find an indicator for these...Ch. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Prob. 19.66PCh. 19 - Write the ion-product expressions for (a) silver...Ch. 19 - Write the ion-product expressions for (a)...Ch. 19 - Write the ion-product expressions for (a) calcium...Ch. 19 - Prob. 19.70PCh. 19 - The solubility of silver carbonate is 0.032 M at...Ch. 19 - Prob. 19.72PCh. 19 - Prob. 19.73PCh. 19 - The solubility of calcium sulfate at 30°C is 0.209...Ch. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Prob. 19.77PCh. 19 - Calculate the molar solubility of Ag2SO4 in (a)...Ch. 19 - Prob. 19.79PCh. 19 - Prob. 19.80PCh. 19 - Prob. 19.81PCh. 19 - Prob. 19.82PCh. 19 - Prob. 19.83PCh. 19 - Write equations to show whether the solubility of...Ch. 19 - Prob. 19.85PCh. 19 - Prob. 19.86PCh. 19 - Prob. 19.87PCh. 19 - Does any solid PbCl2 form when 3.5 mg of NaCl is...Ch. 19 - Prob. 19.89PCh. 19 - Prob. 19.90PCh. 19 - Prob. 19.91PCh. 19 - A 50.0-mL volume of 0.50 M Fe(NO3)3 is mixed with...Ch. 19 - Prob. 19.93PCh. 19 - Prob. 19.94PCh. 19 - Prob. 19.95PCh. 19 - Write a balanced equation for the reaction of in...Ch. 19 - Prob. 19.97PCh. 19 - Prob. 19.98PCh. 19 - Prob. 19.99PCh. 19 - What is [Ag+] when 25.0 mL each of 0.044 M AgNO3...Ch. 19 - Prob. 19.101PCh. 19 - Prob. 19.102PCh. 19 - Prob. 19.103PCh. 19 - When 0.84 g of ZnCl2 is dissolved in 245 mL of...Ch. 19 - When 2.4 g of Co(NO3)2 is dissolved in 0.350 L of...Ch. 19 - Prob. 19.106PCh. 19 - A microbiologist is preparing a medium on which to...Ch. 19 - As an FDA physiologist, you need 0.700 L of formic...Ch. 19 - Tris(hydroxymethyl)aminomethane [(HOCH2)3CNH2],...Ch. 19 - Water flowing through pipes of carbon steel must...Ch. 19 - Gout is caused by an error in metabolism that...Ch. 19 - In the process of cave formation (Section 19.3),...Ch. 19 - Phosphate systems form essential buffers in...Ch. 19 - The solubility of KCl is 3.7 M at 20°C. 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