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Why is the pH of distilled water usually <7? How can you prevent this from happening?

Short Answer

Expert verified

The pH of distilled water usually <7 due to reaction of dissolved carbon dioxide with water.

Step by step solution

01

Concept Used

ph is a measurement of the level of acid or alkali in a substance or liquid.

pH is the negative logarithm of the concentration ofH+ ions in solution.

To calculate the pH of an aqueous solution we need to know the concentration of the Hydronium ion in moles per liter (morality).

Therefore, the pH of a substance is given by:

pH = - log [H3O+].

02

Explanation for pH of distilled water being less than 7.

Carbon dioxide CO2is present in the atmosphere. Thus, dissolved CO2lowers pH of the dissolved water by reacting with it to form carbonic acid. Hence, the concentration of hydronium ion decreases in a solution the pH of distilled water becomes less than 7.

03

Prevention of formation of carbon dioxide.

To prevent the formation of carbonic acid, water can be distilled under an inert atmosphere so that carbon dioxide is not dissolved in water. Most of the dissolvedcarbon dioxide can be eliminated by boiling the distilled water.

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Most popular questions from this chapter

For the reactionHCO3-H++"CO32-, ΔG°=+59.0kJ/molat 298.15K . Find the value of Kfor the reaction.

Given the following equilibria, calculate the concentration of

each zinc species in a solution saturated withZn(OH)2(s)and containing[OH-]

at a fixed concentration of3.2×1027M.

Zn(OH)2(s)Ksp=3×10-16Zn(OH)+β1=1×104Zn(OH)2(aq)β2=2×1010Zn(OH)3-β3=8×1013Zn(OH)42-β3=3×1015

The planet Aragonose (which is made mostly of the mineral

aragonite, or CaCO3) has an atmosphere containing methane and

carbon dioxide, each at a pressure of 0.10 bar. The oceans are

saturated with aragonite and have a concentration of H1equal to

1.8×1027M. Given the following equilibria, calculate how many

grams of calcium are contained in 2.00 L of Aragonose seawater.

CaCO3(s,aragonite)Ca2+(aq)+CO32-(aq)Ksp=6.0×10-9CO2(g)CO2(aq)KCO2CO2)=3.4×10-2CO2(aq)+H2O(l)HCO3-(aq)+H+(aq)K1=4.5×10-7HCO3-(aq)H+(aq)+CO32-(aq)K2=4.7×10-11

Don’t panic! Reverse the first reaction, add all the reactions

together, and see what cancels.

AlthoughKOH,RbOHand CsOHhave little association between metal and hydroxide in aqueous solution,Li+andNa+do form complexes with:

role="math" localid="1663390733265" Li++OH-LiOH(aq)K1=[LiOH(aq)][Li+][OH-]=0.83

Na++OH-NaOH(aq)K1=0.20

Prepare a table like the one in Exercise 6-B showing initial and final Concentrations of Na++OH-NaOH(aq)and (aq)insolution. Calculate the fraction of sodium in the form (aq)at equilibrium.

The equilibrium constant for the reactionH2OH++OH-is1.0×10-14 at25°C . What is the value of K for the reaction 4H2O4H++4OH-?

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