Ask question

Ask question

All categories

3 years ago

11

0 ml of a 0.100 m solution of nh3 is titrated with 0.250m hcl. after 10.0 ml of the hcl has been added, the resultant solution i

s

1 answer:

Eddi Din [679]3 years ago

6 0

I think the solution is 1.00

You might be interested in

Which one of the following would you expect to dissolve in water? limonene sucrose oil atmospheric oxygen none of the above

sdas [7]

-----------sucrose----------------

3 0

3 years ago

1 point If the boy is pushing with 50N of force and the static friction resistance is 70N of force, what will happen? *

solniwko [45]

Nothing, he shouldn’t be able to move it. Think about it like this say you try really hard to push something that is 5,000 pounds and you push as hard as you can. Well you can’t move it bc it weighs more than you can push. I’m sure their is a equation you can use to see how much you can push (body weight=force?)

6 0

3 years ago

Suppose you have just added 100 ml of a solution containing 0.5 mol of acetic acid per liter to 400 ml of 0.5 m naoh. what is th

Tpy6a [65]

$pH = 13.5$

Explanation:

Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

$\text{HAc} + \text{OH}^{-} \to \text{Ac}^{-} + \text{H}_2\text{O}$

The mixture would contain

$0.4 \times 0.5 - 0.1 \times 0.5 = 0.15 \; \text{mol}$ of $\text{OH}^{-}$ and
$0.1 \times 0.5 = 0.05 \; \text{mol}$ of $\text{Ac}^{-}$

if $\text{Ac}^{-}$ undergoes no hydrolysis; the solution is of volume $0.1 + 0.4 = 0.5 \; \text{L}$ after the mixing. The two species would thus be of concentration $0.30 \; \text{mol} \cdot \text{L}^{-1}$ and $0.10 \; \text{mol} \cdot \text{L}^{-1}$ , respectively.

Construct a RICE table for the hydrolysis of $\text{Ac}^{-}$ under a basic aqueous environment (with a negligible hydronium concentration.)

$\begin{array}{cccccccc} \text{R} & \text{Ac}^{-}(aq) &+ & \text{H}_2\text{O}(aq) & \leftrightharpoons & \text{HAc}(aq) & + & \text{OH}^{-} (aq)\\ \text{I} & 0.10 \; \text{M} & & & & & &0.30 \; \text{M}\\ \text{C} & -x \; \text{M}& & & & +x \; \text{M}& & +x \; \text{M} \\ \text{E} & (0.10 - x) \; \text{M} & & & & x \; \text{M} & & (0.30 +x) \; \text{M} \end{array}$

The question supplied the <em>acid</em> dissociation constant $pK_a$ for acetic acid $\text{HAc}$ ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant $pK_b$ for its conjugate base, $\text{Ac}^{-}$ . The following relationship relates the two quantities:

$pK_{b} (\text{Ac}^{-}) = pK_{w} - pK_{a}( \text{HAc})$

... where the water self-ionization constant $pK_w \approx 14$ under standard conditions. Thus $pK_{b} (\text{Ac}^{-}) = 14 - 4.7 = 9.3$ . By the definition of $pK_b$ :

$[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}$

$x \cdot (0.3 + x) / (0.1 - x) = 10^{-9.3}$

$x = 1.67 \times 10^{-10} \; \text{M} \approx 0 \; \text{M}$

$[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}$

$pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5$

6 0

3 years ago

Complete the statement about the environment and its components.

borishaifa [10]

the environment is Healthy

3 0

2 years ago

What do the conclusions tell about the experiments?

Anni [7]

Answer:

the answer would be C of not C is B

4 0

3 years ago

Read 2 more answers