& Havmany grams of NaC H2O must be added to 100 me of .15MH (H2O2
to get a pH 4.44 buffer? 1) Find moles of acid ... leler. 15 M) = .015 mol HC2H2O2 2) Pluginto H H equation ... 4.44= 4.74 + log x=.0075 mol 3) convert to grams..... 0075 ml NaC2H2O2 x 829 -1.0615 q NaleH302
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How to achieve the correct Acid Ratio
n es in th O Direct Method -- add the exact amounts of acid and conjugate baserwis
2) Indirect Boxe Achi patnya beqin my weak acid add strong base until enough acid converted to conjugate base you can entity molos otectes cas. (.) ALEM) .015 mol
HA + OH - A Hoa 4.74+ logos- If you want to learn more check out Who comes to the shepherds to order them to travel to bethlehem?
at 5. Bx
Plug inta # -# envention (x = mol strong bose needed 5.0 x 2.0075... os mora ndirect Acid Addition to begin of weak base + add strong acid until enough base converted to conjugate acid A+ Hyöt -HATH, O
Find moles of base... (.12)6.1510) 5015 mol nät 2 H2O2
(реры, рэв био је још як) и чрк » Н-н арылт боја () . Т. 9 х Don't forget about the age old question of What is the difference between a theory and a theorem?
X- | X-10
4.44-4.74 + laq (015-x
BUFFER EXAMPLES What is pH of butter prepared by adding 15 ml of oolM HCl to 40mL of . ISM NH3 solution? pk of NH3-4.74
NH3 + H+ NH4+ NHa) (40 m.) (.15 M) = 6.0mmol pha=14-4.74 16 mmol 1.5 mmolo Don't forget about the age old question of What are perpendicular lines?
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(15 mL) (IM) = 1.5 mmol
-1.5 -1.5 +15. If you want to learn more check out What is the imposed by the sedition act?
never put the pkp in the equation!
O 1.5 pH=9.26 + log
it simply shows that NHy is a base
1 1.5 mmol
1 pH=9.74 J
* cannot use H-H
equation unless similar
quantities of acid/base...
Helt base is 0 90
What is pt of solution prepared by adding 70 mL of 1 HCl to 40 mL of 15 M NH 3 solution? pk = 4.74
What Niva (NH3) (60m2) ( 15 M) - 6.0 mmol VIDA Find pH this 6 .00 7.0
(H) (70 mc) ( . 1 M.) = 7.0 inmol Find [Ita ut Hequation
(-6.0 16.0 16.0
Immula torovaire ?
40+ TUML 110
you don't always subtract the aid! quantite you subtract the limiting reactant...
-log(.0091 M) = 12.04pH ]
in this case, it is not the strong acid additive!
not a butter &
* only weak acids howe
Ka valves! *
known concentration (needs to be strong to know it
goes to completion)
be acid, one
strong/ weak acid or base
* Titrant is added to analyte to determine the conuentration of the analyte solution.
of unknown concentration
(1) 50 ml of NaOH (analyte) was titrated if 15.26 mL of . I HRI (titrant). What is [NuOH]?
* Equivalence Point / Erd Paint
1) Carenake mood swans... (15.26 mb) (1M) = 1.526 mmol HCl
the number of males of titrant analyte
are the same!
slechinodry te entrale mmel of fayld..... HCI + NAOH → H20+ Nall 1.526 mnd HCI & limmal NaOH = 1.526 mmol Naolt Immo NAOH
* If stochiometry is 1:1,
3) Cala Molenciyle... 1.Szo mmol NaOH -1.03052 M NAOH
Hava = M, V+
somb TYPES of ACID-BASE TITRATIONS O Strong Acid to dissociates 100%.; pH--109[H204]
Strong Buse to dissociates 100% ; poft = -1og [OH-] Weak Aud to partial dissociation; USC CHA), ICE diagram and Ka weak Base + partial dissociation; us [n-J. 1E diagram, and Ky Buffer calculate equilibrium mmol/M of HA+A; use W-4 equation to calculate pH
Ex) Zome of .25M HF (PKG = 317) with at M NaOH.
& Eguvalent volume of titrant... Vea = (20 mL HF) (.25 mmol HELImmol NaOH Im NaOH)
= 50mL NAOH
I mL HF
2 what is pH before any titant added?...
mmol HE olmmal NaOH !
Vitane OmL * HF is only a weak aridu
* HF is only a weak acid when by itself!
titant V total = 20 mL + Ome HF + H2O + H2ot ka=6.76610-4 HF
- log (60127) = 1.89
"ICE table just w analyte and wider.
0127MH + JPH = 1.89
3) pt at 25 mL of titranti (midpoint
titrant = 25 mL
when you mix solutions together, you have to consider moles/mmol.& AF) ZOMLX.25 M=5 mmol HF HF + OH - F + H2O
5 mmol 2.5 mmollo
NASH) 25 mLx. IM2 2.5mmal OH
V total = 20mL + 25 mL
vse molanty when ICE table is
pH - 3,17+10913) -H- 3.17 *pH=pka at the midpoint of tittation 4) pH at 50 mL of futuant lequivalence point) HF+OH +H20 F+H205#F+0# Rp 1.5 x10"
Vritrant - 50 ml
-4714 ZIO 10
total= 20 mL & SomL 70 mL
Loos 1 0714 x
(20 m2) (25) - 5 mmol F
•15+ ICE = arałyle+ tirant is buffer
- Zna ICE switch reactants/prodveta ble not a bute
No longer a butter. Only weak base solution So do artikel table.
x = 1.03 x 10-6