How To Find A Solution With A Ph Of 7 Is

Posted by on May 25, 2012

Thomas asks…

A 22mL solution of 0.21M solution of HF( Ka =7.1 *10^4) is titrated with 0.16M KOH,what is the PH at endpoint?

pls help i need help finding PH the endpoint,

5 points asap.

answers:

Moles HF = 0.022 L x 0.21 M=0.0046
Moles OH- needed = 0.0046
Volume OH- = 0.0046/ 0.16 M=0.029 L
total volume = 0.022 + 0.029=0.051 L

HF + OH- => F- + H2O
concentration F- = 0.0046/ 0.051=0.090 M

F- + H2O HF + OH-
Kb = Kw/Ka = 1.0 x 10^-14/ 7.1 x 10^-4=1.4 x 10^-11 = x^2/ 0.090-x

x = [OH-]= 1.1 x 10^-6 M
pOH = – log 1.1 x 10^-6=5.9
pH = 14 – 5.9=8.1

Ruth asks…

What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79?

using 31.52 g of TrisHCl

I got and answer of 6.7mL
where I got stuck is the second part which is

The buffer from Part A is diluted to 1.00 L. To half of it (500mL), you add 0.0100 mol of hydrogen ions without changing the volume. What is the pH of the final solution?
thanks a lot in advance if anyone can help me out

answers:

Hahah!!! I’m having problems with that too!… MasteringChemistry Sucks!!!

Chris asks…

What is the PH of a solution prepared by dissolving 0.25g of BaO (molar mass = 153.3 g/mol) in enough..?

water to make a 0.5 L solution.

My problem with this question is, why wouldn’t the pH just be 7 since there is no acid here except water, which is neutral. why should the pH here be any higher or lower? Please show all calculations and provide explanations. thank you!!

The correct answer is: 11.81

answers:

You are forgetting that BaO reacts with water according to the following equation:
BaO + H?O —> Ba²? + 2OH?

1) Divide the grams of BaO by the molar mass to figure out the moles of BaO.
0.25g / 153.3g/mol = .0016 moles of BaO

2) For every mole of BaO, there are twice the moles of OH? According to the balanced equation. Multiply the moles of BaO by two.
.0016 * 2 = .0032 moles OH?

3) Divide the moles of OH? By the liters of solution to determine the concentration of OH?.
.0032 moles OH? / 0.5L = .0064M OH?

4) Calculate the pOH by doing -log [OH?]
pOH = -log[OH?]
pOH = -log[.0064]
pOH = 2.19

4) At 25 degrees Celsius, the sum of pOH and pH is 14. Substitute 2.19 for the pOH.
POH + pH = 14
2.19 + pH = 14
pH = 11.81

Lisa asks…

chemistry help!!!!!!!!!!!!!!!!!?

In a titration, 3.00 M NaOH(aq) was added to an erlenmeyer flask containing 25.00 millilliter of HCl(aq) and three drops of phenolphtalein until one drop of the NaOH(aq) turned the solution a light-pink color. The following data were collected by a student performing this titration.

Intial NaOH(aq) buret reading: 14.45 millilliters
final NaOH(aq) buret reading: 32.66 millilliters

what is the total volume of NaOH(aq) that was used in this titration?

do i subtract the Initial from the final?
——————–
a student was studying the pH differences in samples from two adirondack streams. the student measured a pH of 4 in stream A and a pH of 6 in stream B.

what is the color of bromthymol blue in the sample of stream A?

idk
——————–
indicator/color result from the indicator test

methyl orange/yellow
litmus/red

which pH value is consistent with the indicator results?

1
3
5
10

um idk
—————–
how many millilliter of 0.100 M NaOH(aq) would be needed to completely neutralize 50.0 millliliters of 0.300 M HCl(aq)?
16.7 mL
50.0 mL
150. mL
300. mL

dang it i forgot how to do this. i suck in this chemistry math!!!!
——————
in a laboratory activity 0.500 mole of NaOH(s) is completely dissolved in distilled water to form 400. millilliter of NaOH(aq) this soultion is then used to titrate a solution of HNO3(aq).

which answer choice correctly completes the equation by fillling in the formulas of the products in this titration reaction?

NaOH(aq) + HNO3(aq) –> ______+________

NaNO2(aq) + H2O2 (l)
NaNO3(aq) + H2O(l)
HNaNO3(aq) + OH(l)
H2NO3(aq) + NaO(l)

———————–
which volume of 0.10 M naOH(aq) exactly neutralizes 15.0 millilter of 0.20 M HNO3(aq)?
1.5mL
7.5 mL
3.0 mL
30.0 mL

——————–
if 20.0 milliliters of 4.0 M NaOH is exactly neutralized by 20.0 millilliters of HCl, the molarity of the HCl is
1.0 M
2.0 M
5.0M
4.0M
—————-
Given the following solutions:
Solution A: pH of 10
solution b: pH of 7
solution C: pH of 5

which list has the solutions placed in order of increasing H+ concentration?

a,b,c
b,a,c
c,a,b
c,b,a

answers:

In a titration, 3.00 M NaOH(aq) was added to an erlenmeyer flask containing 25.00 millilliter of HCl(aq) and three drops of phenolphtalein until one drop of the NaOH(aq) turned the solution a light-pink color. The following data were collected by a student performing this titration.

Intial NaOH(aq) buret reading: 14.45 millilliters
final NaOH(aq) buret reading: 32.66 millilliters

what is the total volume of NaOH(aq) that was used in this titration?
32.66 – 14.45 =
do i subtract the Initial from the final? Yes
——————–
a student was studying the pH differences in samples from two adirondack streams. The student measured a pH of 4 in stream A and a pH of 6 in stream B.

What is the color of bromthymol blue in the sample of stream A?

Find a pH indicator chart or do a yahoo search for bromthymol blue, and it will tell you the color in acid and base
pH of 4= acid
idk
——————–
indicator/color result from the indicator test

methyl orange/yellow weak acid
litmus/red red = acid

I taught my chemistry students to remember that litmus turned red in acids by holding my tongue and saying, “I have a red apple!” apple sounds lie as_ -ole

which pH value is consistent with the indicator results?
Acids pH ______+________
NaOH ? Na+1 + OH-1
HNO3 ? H+1 + NO3-1
Na+1 + OH-1 + H+1 + NO3-1 ? NaNO3 + H2O

NaNO2(aq) + H2O2 (l)
NaNO3(aq) + H2O(l)
HNaNO3(aq) + OH(l)
H2NO3(aq) + NaO(l)

———————–
which volume of 0.10 M naOH(aq) exactly neutralizes 15.0 millilter of 0.20 M HNO3(aq)?
Normality * Volume = Normality * volume
Moles of acid…. = …Moles of base
0.2 * 15 = 0.1 * Volume
Volume = ____ml of 0.100 M NaOH

1.5mL
7.5 mL
3.0 mL
30.0 mL

——————–
if 20.0 milliliters of 4.0 M NaOH is exactly neutralized by 20.0 millilliters of HCl, the molarity of the HCl is

Moles of acid…. = …Moles of base
Normality * Volume = Normality * volume
M * 20 = 4.0 * 20
Volume = ____ml of 0.100 M NaOH
1.0 M
2.0 M
5.0M
4.0M
—————-
Given the following solutions:
Solution A: pH of 10
solution b: pH of 7
solution C: pH of 5

which list has the solutions placed in order of increasing H+ concentration?

PH < 7 is acid, the lower the pH , the stronger the acid

a,b,c
b,a,c
c,a,b
c,b,a

Linda asks…

Help with calculating a buffer solution?

You wish to prepare 100 mL (TOTAL volume!) of a buffer solution that is .025 M in carbonic acid, pH 7.25. You have solid H2CO3 and solid NaHCO3. Calculate how many grams of H2CO3 and HCO3- you need to prepare this solution.

I’m pretty sure you use henderson-hasselbach, but am still really confused! Thanks!

answers:

7.25=6.36 +log (base/acid)
0.89= log (base/acid)
7.76=base/acid
base=(7.76/8.76) *0.025M/0.1L=0.2214M NaHCO3
acid=(1/8.76) * 0.025M/0.1L=0.0285M H2CO3
Need 0.1L*0.025M NaHCO3= 0.0025 moles NaHCO3
Convert that number of moles of NaHCO3 to grams and you should be good to go. (0.0025 * molar mass of NaHCO3)

Check my math on all this, we’re probably taking the same level class.

Powered by Yahoo! Answers

Leave a Reply

Your email address will not be published. Required fields are marked *

*

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>