In Example 16-9, rather than use the quadratic formul at o solve the quadratic c equation, we could have proceeded in the following way Substitute the value yielded by our failed assumption — x = 0.0010—into the denominator of the quadrati c equation: that is, use 10 00250 -0.00102 as the value of [CH 2 NH 2 ] and solve for a new value of x Use this second value of x to re-evaluate [CH 3 NH 3 ]: [CH 2 NH 3 ] = 10 00250 -second value of x2 Solve the simple quadratic equation far a third value of x and so on. After three or four tnals. you will find that the value of x no longer changes This is thanswer you are seeking (a) Complete the calculation of the pH of 0.0250 M CH 2 NH 2 by this method and show that the result is the same as that obtained by using the quadratic formula (b) Use this method to determine the pH of 0.500 M HCIO
In Example 16-9, rather than use the quadratic formul at o solve the quadratic c equation, we could have proceeded in the following way Substitute the value yielded by our failed assumption — x = 0.0010—into the denominator of the quadrati c equation: that is, use 10 00250 -0.00102 as the value of [CH 2 NH 2 ] and solve for a new value of x Use this second value of x to re-evaluate [CH 3 NH 3 ]: [CH 2 NH 3 ] = 10 00250 -second value of x2 Solve the simple quadratic equation far a third value of x and so on. After three or four tnals. you will find that the value of x no longer changes This is thanswer you are seeking (a) Complete the calculation of the pH of 0.0250 M CH 2 NH 2 by this method and show that the result is the same as that obtained by using the quadratic formula (b) Use this method to determine the pH of 0.500 M HCIO
Solution Summary: The author explains the ionization of a base B by comparing it with the result obtained by quadratic equation method.
In Example 16-9, rather than use the quadratic formul at o solve the quadratic c equation, we could have proceeded in the following way Substitute the value yielded by our failed assumption — x = 0.0010—into the denominator of the quadrati c equation: that is, use 10 00250 -0.00102 as the value of [CH2NH2 ] and solve for a new value of x Use this second value of x to re-evaluate [CH3NH3]: [CH2NH3] = 10 00250 -second value of x2 Solve the simple quadratic equation far a third value of x and so on. After three or four tnals. you will find that the value of x no longer changes This is thanswer you are seeking (a) Complete the calculation of the pH of 0.0250 M CH2NH2 by this method and show that the result is the same as that obtained by using the quadratic formula (b) Use this method to determine the pH of 0.500 M HCIO
The chemical reaction you investigated is a two-step reaction. What type of reaction occurs in each step? How did you determine your answer?
What is the relationship between the limiting reactant and theoretical yield of CO2?
From your calculations, which reaction experiment had closest to stoichiometric quantities? How many moles of NaHCO3 and HC2H3O2 were present in this reaction?
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