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Subject Test Chemistry

Question 8 of 17

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SAT Subject Test

Each of the questions or incomplete statements below is followed by five suggested answers or completions.  Select the one that is best in each case.

A solution is made by adding 5.6 invisible times text g end text of text KOH end text (molar mass 56 invisible times fraction numerator text end text text g end text over denominator text mol end text end fraction) to enough water to make 1.0 invisible times text L end text of solution. What is the approximate text pH end text of the resulting solution?

Select an Answer

1

Correct Answer: 
No

3

Correct Answer: 
No

7

Correct Answer: 
No

9

Correct Answer: 
No

13

Correct Answer: 
Yes

The 5.6 grams of text KOH end text is 0.1 mole of text KOH end text. A solution of 0.1 mole of solute in 1 liter of solution has a concentration of 0.1 invisible times text      end text Mtext KOH end text is a strong base, and dissociates completely in water. Therefore, the concentration of text OH end text to the power of minus in this solution is 0.1 invisible times text     end text M.

left square bracket text OH end text to the power of minus end exponent right square bracket invisible times equals 1 invisible times cross times 10 to the power of negative 1 end exponent

K subscript w end subscript invisible times equals left square bracket text H end text to the power of plus end exponent right square bracket left square bracket text OH end text to the power of minus end exponent right square bracket invisible times equals 1 invisible times cross times 10 to the power of negative 14 end exponent

left square bracket text H end text to the power of plus end exponent right square bracket invisible times equals fraction numerator left parenthesis 1 invisible times cross times 10 to the power of negative 14 end exponent right parenthesis over denominator left parenthesis 1 invisible times cross times 10 to the power of negative 1 end exponent right parenthesis end fraction invisible times equals 1 invisible times cross times 10 to the power of negative 13 end exponent  

text pH = - end text text log end text left square bracket text end text text H end text to the power of plus end exponent right square bracket, so text pH = 13 end text  

Question Difficulty: 
medium