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saw5 [17]
3 years ago
13

Suppose you prepare a buffer by adding 0.157 mol of NaH2PO4 and 0.966 mol of Na2HPO4 to 1.0 L of water. What is the pH of the so

lution?
Chemistry
1 answer:
Arlecino [84]3 years ago
5 0

Answer:

pH = 7,0

Explanation:

The buffer producing by the mixture of H₂PO₄⁻ and HPO₄²⁻ is:

H₂PO₄⁻ ⇄ HPO₄²⁻ + H⁺; pka: 7,21

Using the Henderson-Hasselbalch formula:

pH = pka + log₁₀ A⁻ / HA

Where: A⁻ is the conjugate base (HPO₄²⁻) and HA is the weak acid (H₂PO₄⁻)

Replacing the formula:

pH = 7,21 + log₁₀ HPO₄²⁻ / H₂PO₄⁻

As the moles of HPO₄²⁻ are 0,0966 and moles of H₂PO₄⁻ are 0,157:

pH = 7,21 + log₁₀  0,0966 / 0,157

pH = 7,0

I hope it helps!

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P = nRT/V

P = 3.5 x 10^-3 x 0.082 x 298 /0.5

P = 0.171 m Hg

P = 171 mm Hg

hope this helps
8 0
3 years ago
2 Nobr +heat > 2 no2+br2 what happens when you remove nobr
Kryger [21]
There would be no nobr and it would just be 2
6 0
3 years ago
How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?
givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

H2SO3\leftrightarrow H^{+}+HSO3^{-} --------(1)

where,  Ka1 = 1.5 x 10–2  or pKa1 = 1.824

HSO3^{-}\leftrightarrow H^{+}+SO3^{2-} --------(2)

where,  Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

i.e. \ 5.13\  moles/L*0.385L = moles\  NaOH\\therefore, \ moles\  NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L

Step 2:

For the second equivalence point setup an ICE table:

                  HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}

Initial           1.98                    ?                                       0

Change      -x                       -x                                       x

Equil           1.98-x                 ?-x                                    x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles

Volume of NaOH required is:

\frac{0.634\ moles}{0.615 moles/L}=0.389L

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

3 0
3 years ago
A solution containing 42% naoh by mass has a density of 1.30 g/ml. what mass, in kilograms, of naoh is in 6.00 l of this solutio
Aliun [14]
NaOH
Na=23
O=16
H=1
SO,NaOH=23+16+1
NaOH=40g/mol
42% of 40=42/100x40
oxygen is the compound
density=mass/volume
1.30=mass/6
mass=7.8g
3 0
4 years ago
Read 2 more answers
Which statement best describes the effect of beta particles on body tissue?
mash [69]
I think the answer is C
3 0
3 years ago
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