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astraxan [27]
3 years ago
12

A chemist prepares a solution of calcium sulfate (CaSO_4) by measuring out 0.0101 g of calcium sulfate into a 350. mL volumetric

flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's calcium sulfate solution. Round your answer to 3 significant digits.
Chemistry
1 answer:
Citrus2011 [14]3 years ago
3 0

Answer:

2.123x10⁻⁴ M (mol/L)

Explanation:

First, we are considering the molar mass for each element, a data we obtain from the periodic table:

Ca= 40

S= 32

O= 16

Then, for CaSO₄, the molecular mass is 136 g per mol

As we dissolve 0.00101 g of this salt, we are going to convert this mass in moles units:

1 mol CaSO₄=136 g → 0.0101 g=7.43x10⁻⁵ mol

We finally obtain the concentration for the 350 ml, having in mind that it should be in mol/L:

350ml=7.43x10⁻⁵ mol

1000ml (1L)=2.123x10⁻⁴ M , where M=mol/L

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Answer:

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snow_lady [41]

Answer:

\displaystyle \text{p} K_a \approx 3.856

Explanation:

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\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right]  & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}

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\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:

\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}

In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.

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