0.40 L of HNO3 is titrated to equivalence using 0.18 L of 0.1 M NaOH. What is the concentration of the HNO3 ?

1 answer:

3 0

This problem is providing us with the volume of nitric acid that is titrated with 0.18 L of 0.1-M sodium hydroxide and asks for the concentration of the acid. At the end, the result turns out to be 0.045M, according to the following.

<h3>Acid-base titrations:</h3>

In chemistry, acid-base titrations allow us to quantify the volume or concentration of an acid or base via the following equation:

$M_AV_A=M_BV_B$

Where the subscript A stands for the acid and B for the base; which means one can calculate any of the variables there by knowing the other three. This equation is based on the balanced neutralization chemical equation, which takes place between the acid and the base.

Thus, we can write the reaction between NaOH and HNO3 as:

$HNO_3+NaOH\rightarrow NaNO_3+H_2O$

In such a way, we can solve for the concentration of the acid as shown below:

$M_A=\frac{M_BV_B}{V_A} \\\\M_A=\frac{0.1M*0.18L}{0.40L} \\\\M_A=0.045M$

Learn more about titration: brainly.com/question/25485091

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140000 rounded to 3 sig gigs

attashe74 [19]

1.40E+5 (140000)
I believe this is right

8 0

3 years ago

3. If you start with 8x1025 molecules of Cl, and 25 grams of KI, how many grams of KCl would

miskamm [114]

Answer:

Percent yield = 89.1%

Explanation:

Based on the equation:

Cl₂ + 2KI → 2KCl + I₂

<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>

<em />

To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:

<em>Moles Cl₂:</em>

8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles

<em>Moles KI -Molar mass: 166.0028g/mol-</em>

25g * (1mol / 166.0028g) = 0.15 moles

Here, clarely, the KI is the limiting reactant

As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:

0.15 moles * (74.5513g / mol) =

11.2g KCl

Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100

<h3>Percent yield = 89.1%</h3>

3 0

3 years ago