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Methyl orange is an indicator with a Ka of 10^(-4). Its acid form HIn is red, while its base form In- is yellow. At pH 6.
Methyl orange is an indicator with a Ka of 10^(-4). Its acid form HIn is red, while its base form In- is yellow.At pH 6.0, the indicator will be (yellow <= correct answer)I don't understand why this is the correct answer. Here is my understanding, please correct me where I am incorrectWe know that - log Ka = pKa.Therefore, - log (10^(-4)) = 4.Therefore pKa =4.pH = pKa at half-equivalence point, so at half equivalence pt, the pH is four.At equivalence point, we have the same amount of acid and base). Therefore, 2 * half equivalence point = equivalence point. For our problem, 2*4=8, therefore 8 is our equivalence point. So, at a pH of 6, a pH of 6 < a pH of 8, therefore, according to my logic, the indicator would be red because we would not have reached equivalence point yet, and the solution would still be acidic. But my logic is obviously incorrect, so please help me find out where I am wrong. Thanks.
Methyl orange is an indicator with a Ka of 10^(-4). Its acid form HIn is red, while its baseform In- is yellow.At pH 6.0, the indicator will be (yellow <= correct answer)I don't understand...