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stealth61 [152]
3 years ago
6

In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?

Chemistry
2 answers:
Rom4ik [11]3 years ago
8 0
<span>M(HCl) * </span><span>V(HCl) </span>= <span>M(NaOH) * </span><span>V(<span>NaO<span>H)
</span></span></span>
M(HCl) = 0.35
<span>V(HCl) = 45mL
</span>M(NaOH)= 0.35

now, solne for V(NaOH) by putting these values in the above equation.
M(HCl) * <span>V(HCl) </span>= <span>M(NaOH) * </span><span>V(NaOH)</span>

<span>0.35 * 45 = 0.35 * V(NaOH)</span>

<span>V(NaOH) = 45 mL</span>

stich3 [128]3 years ago
6 0

how it cumz is

acc to the law of volumetric anlysis

V1N1 = V2N2

take V1 as vol of HCl = 0.35 M

V2 as vol of NaOH = 0.35M

N1 of HCl = 45.0

N2 of NaOH = ?

N2 = V1 x N1

V2

= 0.35 x 45.0

----------------

0.35

= 45.0

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3 years ago
What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3
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Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:

V acid *M acid = V base *M base

where V represents the volume of solution and M the molar concentration of said solution.

In this case:

  • V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
  • M acid= 0.129 M
  • V base= ?
  • M base= 0.135 M

Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

Solving:

V base=\frac{0.0137 L*0.129 M}{0.135 M}

V base=0.0131 L = 13.1 mL

<u><em> The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>

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A sample of lead has a density of 11.3 g/cm3 and a mass of 123 grams, what is the volume of the sample?
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Answer:

The answer is

<h2>11.18 cm³</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question

mass = 123 g

density = 11.3 g/cm³

The volume is

volume =  \frac{123}{11.3}   \\  = 11.18181818...

We have the final answer as

<h3>11.18 cm³</h3>

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