Question Distilled white vinegar is used for cooking but can also be used in many applications for cleaning. The active ingredient in vinegar is acetic acid (CH3COOH), which has a pKa= 4.74). If household vinegar produces a pH of 4.2,what is the initial concentration(in M)of CH 3 3 ​ COOH (assuming it is the only substance contributing to pH)? Question 1 Homework. Answered Distilled white vinegar is used for cooking but can also be used in many applications for cleaning. The active ingredient in vinegar is acetic acid (CH3COOH), which has a pka=4.74). If household vinegar produces a pH of 4.2,what is the initial concentration(in M)of CH3COOH (assuming it is the only substance contributing to pH)? O A 2.19x10-4 0 B 0.10 o C 4.10 x 103 OD 0.87

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Distilled white vinegar is used for cooking but can also be used in many applications for cleaning. The active ingredient in vinegar is acetic acid (CH3COOH), which has a pKa= 4.74). If household vinegar produces a pH of 4.2,what is the initial concentration(in M)of CH 3 3 ​ COOH (assuming it is the only substance contributing to pH)?

Transcribed Image Text: Question 1 Homework. Answered Distilled white vinegar is used for cooking but can also be used in many applications for cleaning. The active ingredient in vinegar is acetic acid (CH3COOH), which has a pka=4.74). If household vinegar produces a pH of 4.2,what is the initial concentration(in M)of CH3COOH (assuming it is the only substance contributing to pH)? O A 2.19x10-4 0 B 0.10 o C 4.10 x 103 OD 0.87
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Transcribed Image Text: Question 1 Homework. Answered Distilled white vinegar is used for cooking but can also be used in many applications for cleaning. The active ingredient in vinegar is acetic acid (CH3COOH), which has a pka=4.74). If household vinegar produces a pH of 4.2,what is the initial concentration(in M)of CH3COOH (assuming it is the only substance contributing to pH)? O A 2.19x10-4 0 B 0.10 o C 4.10 x 103 OD 0.87
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    Answer :   The initial concentration of acetic acid is A). 2.19 × 10^-4 M     Explanation :   Here given data :   • pH of solution = 4.2   • pKa of acid = 4.74   Step 1 of 3   Calculate [H+] from pH :           We know that,   pH = - log [H+]   => [H+] = 10^-pH      => [H+] = 10^- 4.2   => [H+] = 6.31 × 10^-5  M   Step 2 of 3   Calculate Ka from pKa :     pKa = - ... See the full answer