Write a question you have about Titan or about methane that could help you determine what happened to the lake

Chemistry

1 answer:

Sergeeva-Olga [200]3 years ago

8 0

Answer: what would swimming in mathane lake on titan feel like.

Explanation:

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Use the Henderson-Hasselbalch equation, eq. (3), to calculate the pH expected for a buffer solution prepared from this acid and

notka56 [123]

Answer:

$pH=4.56$

Explanation:

Hello there!

In this case, given the Henderson-Hasselbach equation, it is possible for us to compute the pH by firstly computing the concentration of the acid and the conjugate base; for this purpose we assume that the volume of the total solution is 0.025 L and the molar mass of the sodium base is 234 - 1 + 23 = 256 g/mol as one H is replaced by the Na:

$n_{acid}=\frac{0.2g}{234g/mol}=0.000855mol\\\\n_{base}= \frac{0.2g}{256g/mol}=0.000781mol$

And the concentrations are:

$[acid]=0.000855mol/0.025L=0.0342M$

$[base]=0.000781mol/0.025L=0.0312M$

Then, considering that the Ka of this acid is 2.5x10⁻⁵, we obtain for the pH:

$pH=-log(2.5x10^{-5})+log(\frac{0.0312M}{0.0342M} )\\\\pH=4.60-0.04\\\\pH=4.56$

Best regards!

6 0

2 years ago

What is the ph of a solution of 0.550 m k2hpo4, potassium hydrogen phosphate?

Solnce55 [7]

We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:

PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32

3 0

3 years ago

(GIVING BRAINLIEST) do what the following image says.

S_A_V [24]

Answer:

hope it helps.

<h3>stay safe healthy and happy.<u>.</u><u>.</u></h3>