Concentration of KHP Solution 12753 gramsKHP 1 x 1 molKHP 20422 KHP 0062447

Concentration of khp solution 12753 gramskhp 1 x 1

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Concentration of KHP Solution: ( 12.753 gramsKHP 1 ) x ( 1 molKHP 204.22 KHP ) = 0.062447 molKHP M KHP = 0.062447 mol 0.250 L
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= 0.250M KHP Concentration of the Standardized NaOH Solution: Chemical Equation : NaOH (aq ) + KHP (aq) H 2 O (l) + NaKP (aq) Trial 1 : (0.01413liters)(M NaOH ) = (0.01501liters)(0.250M KHP ) → M NaOH = ( 0.01501 liters )( 0.250 M KHP ) ( 0.01413 liters ) = 0.266M NaOH Trial 2: (0.01387liters) ( M NaOH ) = (0.01501liters)(0.250M KHP ) → M NaOH = ( 0.01501 liters )( 0.250 M KHP ) ( 0.01387 liters ) = 0.271M NaOH Trial 3: (0.01430liters)(M NaOH ) = (0.01501liters)(0.250M KHP ) → M NaOH = ( 0.01501 liters )( 0.250 M KHP ) ( 0.01430 liters ) = 0.262M NaOH Average Concentration : ( 0.266 M NaOH + 0.271 M NaOH + 0.262 M NaOH ) 3 = 0.266M NaOH Concentration of the Standardized HCl Solution: Chemical Equation :NaOH (aq) + HCl (aq) H 2 O (aq) + NaCl (aq) Trial 1: (0.01491liters)(0.266M NaOH ) = (0.01501liters)(M HCl ) → M HCl = ( 0.01491 liters )( 0.266 M NaOH ) ( 0.01501 liters ) = 0.264M HCl Trial 2: (0.01508liters)(0.266M NaOH ) = (0.01501liters)(M HCl ) → M HCl = ( 0.01508 liters )( 0.266 M NaOH ) ( 0.01501 liters ) = 0.267M HCl Trial 3: (0.01520liters)(0.266M NaOH ) = (0.01501liters)(M HCl ) → M HCl = ( 0.01520 liters )( 0.266 M NaOH ) ¿ ¿ ( 0.01501 liters ) = 0.269M HCl Average Concentration : ( 0.264 M HCl + 0.267 M HCl + 0.269 M HCl ) 3 = 0.267M HCl Concentration of the Standardized Unknown HCl Solution:
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Chemical Equation: NaOH (aq) + HCl (aq) H 2 O (aq) + NaCl (aq) Trial 1: (0.01669liters)(0.266M NaOH ) = (0.01501liters)(M HCl ) → M HCl = ( 0.01669 liters )( 0.266 M NaOH ) ( 0.01501 liters ) = 0.296M HCl Trial 2: (0.01649liters)(0.266M NaOH ) = (0.01501liters)(M HCl ) → M HCl = ( 0.01649 liters )( 0.266 M NaOH ) ( 0.01501 liters ) = 0.292M HCl Trial 3: (0.01679liters)(0.266M NaOH ) = (0.01501liters)(M HCl ) → M HCl = ( 0.01679 liters )( 0.266 M NaOH ) ( 0.01501 liters ) = 0.298M HCl Trial 4: (0.01674liters)(0.266M NaOH ) = (0.01501liters)(M HCl ) → M HCl = ( 0.01674 liters )( 0.266 M NaOH ) ( 0.01501 liters ) = 0.297M HCl Average Concentration : ( 0.296 M HCl + 0.292 M HCl + 0.298 M HCl + 0.297 MHCl ) 4 = 0.296M HCl Discussion Questions: 1. When you titrate an acid with a base, you are advised to titrate to the faintest pink color. Why is this done? The reason why we are advised to titrate to the faintest pink color is because the faintest pink color tells us that we have reached the end point and are very close to the equivalence point. Having a darker pink color tells us that we have passed the equivalence point and that there is excess titrate in the titrand. The faintest pink color is much more accurate because it tells us that there is very little excess titrate, which is what we do not want. 2. Discuss the titration process - was your titration successful? Was your data precise? The titrands, 15.01mL of KHP, 15.01mL of HCl, and 15.01 mL of unknown HCl, were placed in a 250 mL Erlenmeyer flask along with two drops of phenolphthalein indicator . Then, the NaOH
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was added into the buret and dripped into the titrand until the endpoint, when the solution changed to a very faint pink color, was reached. Since the endpoint is approximately the point where the equivalence point is reached, the more precise the titration would be. Our data was indeed precise since each titration resulted in a very faint pink color. Thus, each measurement should be close to the actual concentration of the solution. Even though a slight area of error is present, the data provided should still be precise given that the endpoint of each titration was titrated accurately.
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