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Mandarinka [93]
3 years ago
15

Calculate the concentration of H3O⁺ in a solution that contains 5.5 × 10-5 M OH⁻ at 25°C. Identify the solution as acidic, basic

, or neutral. A) 1.8 × 10-10 M, basic B) 1.8 × 10-10 M, acidic C) 5.5 × 10-10 M, neutral D) 9.2 × 10-1 M, acidic E) 9.2 × 10-1 M, basic
Chemistry
1 answer:
harina [27]3 years ago
3 0

Answer : The correct option is, (A) 1.8\times 10^{-10}M, basic.

Explanation : Given,

Concentration of OH^- ion = 5.5\times 10^{-5}M

First we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (5.5\times 10^{-5})

pOH=4.26

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.26=9.74

Now we have to calculate the H_3O^+ concentration.

pH=-\log [H_3O^+]

9.74=-\log [H_3O^+]

[H_3O^+]=1.8\times 10^{-10}M

As we know that, when the pH value is less than 7 then the solution acidic in nature and when the pH value is more than 7 then the solution basic in nature.

From the pH value, 9.74 we conclude that the solution is basic in nature because the value of pH is greater than 7.

Therefore, the H_3O^+ concentration is, 1.8\times 10^{-10}M, basic.

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What is the percent yield for a process in which 10.4g of CH3OH reacts and 10.1 g of CO2 is formed
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Answer:

A. 70.7%

Explanation:

In the first step lets compute the molar mass of CH₃OH and CO

Molar Mass of CH₃OH =  1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)

                                     = 32.042 g/mol

Molar Mass of CO₂      = 1(12.01 g/mol) + 2(16.00 g/mol)  

                                     = 44.01 g/mol

                                   

Mass of only one reactant i.e. CH₃OH is given so  it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:

Given mass of CH₃OH is 10.4 g. So we have:

                                     10.4g CH₃OH

Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:

                          1 mol CH₃OH / 32.042 g CH₃OH

Convert CH₃OH to moles of CO₂ using mole ratio as:

                             2 mol CO₂ / 2 mol CH₃OH

Convert moles of  CO₂ to grams of  CO₂ utilizing molar mass of  CO₂ as:

                           44.01 g/mol CO₂ / 1 mol CO₂

Now calculating theoretical yield using above steps:

[ 10.4 g CH₃OH ]  [1 mol CH₃OH / 32.042 g CH₃OH ]  [2 mol CO₂ / 2 mol CH₃OH]  [44.01 g/mol CO₂ / 1 mol CO₂]

Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:

= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042

= 457.704/32/042

=  14.28 g CO₂

Theoretical yield =  14.28 g CO₂  

Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:

percent yield = (actual yield / theoretical yield) x 100

As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,

percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%

                       = 0.707 x 100%

                       = 70.7 %

Hence option A 70.7% yield is the correct answer.

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