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gayaneshka [121]
2 years ago
5

Calculation: If you have a pH of 5.5 for a weak acid with a pKa of 4.76, then is there more A- or more HA in the solution? Expla

in why in words using your knowledge of positive or negative log numbers.
Chemistry
2 answers:
devlian [24]2 years ago
6 0

Answer:

lets set the ratio -A/HA as R:

pH = pKa + log(R,10) => pKa + log10(R)

pH = 5.5

pKa = 4.76

R => 10^(pH - 4.76)

10^(pH - 4.76) => 5.4954

Given R (-A/HA) a number bigger than 1, then the concentration of  -A is bigger than HA

Explanation:

meriva2 years ago
6 0

Answer:

There is more A⁻ than HA in the solution  

Explanation:

The equation for the ionization of a weak acid is

HA + H₂O ⇌H₃O⁺ + OH⁻

When HA and A⁻ are present in comparable amounts, we can use the Henderson-Hasselbalch equation:

\begin{array}{rcl}\text{pH} &=& \text{pK}_{\text{a}} + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\5.5 & = & 4.76 + \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\0.74 & = & \log\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\end{array}

0.74 is a positive number, and a number must be greater than one for its logarithm to be positive. That is,

\begin{array}{rcl}\dfrac{[\text{A}^{-}]}{\text{[HA]}} & > & 1\\\\\textbf{[A]}^{-} & > & \textbf{[HA]}\\\end{array}

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HI(aq)+NaOH(aq)→ <br> what the final balanced chemical equation with the phases included
julia-pushkina [17]

Answer:

HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)

Explanation:

Hello there!

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HI(aq)+NaOH(aq)\rightarrow NaI+H_2O

Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:

HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)

Which is also balanced as the number of atoms of all the elements is the same at both sides.

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7 0
3 years ago
A pycnometer is a precisely weighted vessel that is used for highly accurate density determinations. Suppose that a pycnometer h
dexar [7]

Answer:

5.758  is the density of the metal ingot in grams per cubic centimeter.

Explanation:

1) Mass of pycnometer = M = 27.60 g

Mass of pycnometer with water ,m= 45.65 g

Density of water at 20 °C = d =998.2 kg/m^3

1 kg = 1000 g

1 m^3=10^6 cm^3

998.2 kg/m^3=\frac{998.2 \times 1000 g}{10^6 cm^3}=0.9982 g/cm^3

Mass of water ,m'= m - M = 45.65 g -  27.60 g =18.05 g

Volume of pycnometer = Volume of water present in it = V

Density=\frac{Mass}{Volume}

V=\frac{m'}{d}=\frac{18.05 g}{0.9982 g/cm^3}=18.08 cm^3

2) Mass of metal , water and pycnometer = 56.83 g

Mass of metal,M' =  9.5 g

Mass of water when metal and water are together ,m''= 56.83 g - M'- M

56.83 g - 9.5 g - 27.60 g = 19.7 g

Volume of water when metal and water are together = v

v=\frac{m''}{d}=\frac{19.7 g}{0.9982 g/cm^3}=19.73 cm^3

Density of metal = d'

Volume of metal = v' =\frac{M'}{d'}

Difference in volume will give volume of metal ingot.

v' = v - V

v'=19.73 cm^3-18.08 cm^3=

v'=1.65 cm^3

Since volume cannot be in negative .

Density of the metal =d'

=d'=\frac{M'}{v'}=\frac{9.5 g}{1.65 cm^3}=5.758 g/cm^3

5 0
3 years ago
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