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andrezito [222]
3 years ago
10

er solutions can be produced by mixing a weak acid with its conjugate base or by mixing a weak base with its conjugate acid. The

Henderson-Hasselbalch equation, pH=pKa+log[base][acid] allows you to calculate the pH of a buffer. Note that molarity, moles, and millimoles are all proportional, so you can substitute the number moles, or millimoles, for the concentration terms in this formula. Part A You need to produce a buffer solution that has a pH of 5.03. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74.
Chemistry
1 answer:
liubo4ka [24]3 years ago
7 0

Answer:

You need to add 19,5 mmol of acetates

Explanation:

Using the Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [base]/[acid]

For the buffer of acetates:

pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]

As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:

5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]

1,95 = [CH₃COO⁻]/[10]

<em>[CH₃COO⁻] = 19,5 milimoles</em>

Thus, to produce an acetate buffer of 5,03 having 10 mmol of acetic acid, you need to add 19,5 mmol of acetates.

I hope it helps!

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By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling
vova2212 [387]

It is found that scuba diver’s can safely ascend up to 52.53 ft without Breathing out.

By what factor would a scuba diver’s lungs expand if she ascended rapidly to the surface from a depth of 125 ft without inhaling or exhaling then

p_{1} =patm+pH_{2} O

p_{1} = 101325+ρgh_{1}

It is given that height is 125ft. Put the value of h in above formula:

h1  =125ft=38.1m

ρ=1.04g/mL=1040kg/m^{3}

g=9.81                                  

p_{1} =101325Pa+388711.44

p_{1}   =490036.44‬Pa                

p_{2} =p atm  =101325Pa

It is known that volume and pressure can be expressed as:

V*P=const.

where, V is volume and P is pressure.

Now,              

V_{1} *p_{1} =V_{2}  *p_{2}

V_{2} /V_{1}=p_{2} /p_{1}

V_{2} /V_{1} =490036.44/101325

V_{2} /V_{1}=4.84

Assume constant temperature

d of seawater = 1.04 g/mL; d of Hg = 13.5 g/mL.

now p_{1} =p atm​+pH_{2} O =490036.44Pa

V*p=const                              

where, V is volume and P is pressure.

Now,              

V_{1} *p_{1} =V_{2}  *p_{2}

V_{2} /V_{1}=p_{2} /p_{1}

V_{2} /V_{1} =490036.44/X

p_{2} =490036.44pa/(V2/V1) =326690.96Pa

p_{2} =patm +pH_{2} O

p_{2} =101325Pa+ρgh_{2}

326690.96Pa=101325Pa+ρgh_{2}

ρgh1  =151987.5-101325=225365.96‬‬Pa

ρ=1,04g/mL=1040kg/m3

g=9.81

h_{2} =225365.96/‬ρ∗g

​h_{2}  =225365.96  / ‬1040∗9.81

h_{2}  =22.09m= 72.47ft

ΔH=H_{1} -H_{2}

=125-72.47

=52.53ft

So she can safely ascend up to 52.53 ft without Breathing out

To know more about Scuba diver here

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6 0
1 year ago
Suppose you used 0.5 M NaOH to titrate your vinegar sample instead of 0.1 M. What effect does the concentration of base added ha
laiz [17]

Answer: the reliability will be worse

Explanation:

Suppose we used 0.5 M NaOH to titrate our vinegar sample instead of 0.1 M.

Now by using 0.5M instead of 0.1M we are increasing the concentration of NaOH,

We know that  the moles used = Volume x concnetration.

so for the same no of moles, if the concentration increases, the volume decreases.

Hence it will consume less NaOH.

now Since the volume decreases, the titration volume of less number will increase the % error.

Therefore the reliability will be worse.      

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2 years ago
Which of the following has the greatest influence on the size and force of waves?
svlad2 [7]
Wind speed bc The faster the wind, the longer it blows, or the farther it can blow uninterrupted, the bigger the waves.
3 0
2 years ago
Read 2 more answers
At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
2 years ago
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