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CHEM 110 Lecture Notes - Lecture 1: Conjugate Acid, Sodium Hydroxide, Stoichiometry

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30 Jun 2020

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Determining the ph of the contents in a flask in a titration scenario (ignoring any effect of water) strong acid analyte / strong base titrant. Hx + moh h2o + mx (neutral salt) After equivalence point weak acid analyte / strong base titrant. Ha + oh- h2o + a- (conjugate base) After equivalence point weak base analyte / strong acid titrant. B + hx x- + hb+ (conjugate acid) Stoichiometry calculate the remaining [hx], then ph = -log[h+] Kb = x2/[a-] poh = -log x ph = 14 - poh. Determine excess [oh-], then ph = 14 - poh. Kb = x2/[b] poh = -log x ph = 14 - poh. Determine excess [hx], then ph = -log [h+] Note the following: the ph = 7 at the equivalence point for the titration of a strong acid with a strong base. Hpr = a weak acid, so pr- is its conjugate base.

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Related Questions

pH operates off of a ___________ scale.

The autoionization constant of water is abbreviated ________.

The molar concentration of hydronium in pure water at 25ÂºC has been measured to be ________.

The relationship between hydronium and hydroxide in any dilute aqueous solution is represented by _________.

The p in pH stands for _________.

The relationship between pH and pOH is ___________.

The pH value of a 1 M solution of strong acid would be _________.

Neutral pH is ________.

An indicator is typically a ____________ acid or base.

The pOH value of a 1 M solution of strong acid would be _________.

Choose the answer below:

A. Ka = [H3O+][OH-]

B. 1x10-14

C. Kw = [H3O+][OH-]

D. 14

E. - log

F. Ka

G. 1.0x10-7 M

H. Kw

I. log

J. 7

K. 0

L. strong

M. natural log

N. 14 = pH + pOH

O. weak

please answer all parts!! running out of time!

a) You will be titrating a 1.50 M solution of HCN. What is the initial pH of this solution before the titration has begun? b) We measure out 21.00 mL of a 1.50 M solution of HCN for our titration with sodium hydroxide. If you add 1.00 mL of a 0.30 M solution of sodium hydroxide, what is the pH of the acid solution at this stage of the titration? The Ka for HCN is 4.9 x 10-10 1.00 mL of sodium hydroxide added to the weak acid 0.30M sodium hydroxide in the buret as the titrant 21.0 mL of original 1.50 M HCN sample c) If we add enough sodium hydroxide to neutralize 1/2 of our HCN, we will have converted 1/2 the HCN into CN. At this point the [HCN] = [CNT, what is the pH at this point that we call the half equivalence point? d) If we add enough sodium hydroxide to neutralize all of the HCN, we will convert it all into CN and we will have reached the equlvalence point. Since the only thing that remains at the equivalence point is the salt of the conjugate base, CN, the pH at equence will be determined by the hydrolysis of this salt. Since you know how many moles of HCN you started with, you know how many moles of CN must be present at the equivalence point. To calculate the [CN ] you must know the total volume at the equivalence point. How many mL of 0.30 M sodium hydroxide are needed to reach the endpoint? e) What is the pH at the equivalence point of this titration? Remember that since the conjugate base, CN, is hydrolyzing in solution, you will need a Kb value for this base f) If we pass the equivalence point then the pH of the solution is dominated by the strong base. It is like a limiting reagent problem where now, we have run out of the weak acid and there is excess strong base. Even though there will be CN in solution, the OH from the strong base will control the pH. Thus, it is like a strong base pH problem. If we add 115.00 mL of sodium hydroxide we will be well past the equivalence point of the titration. What will be the pH at this point. (Hint: how much strong base will remain unreacted in solution?)