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Alexxandr [17]
3 years ago
6

A solution of hydrochloric acid of unknown concentration was titrated with 0.10 M NaOH. If a 100.-mL sample of the HCl solution

required exactly 1.0 mL of the NaOH solution to reach the equivalence point, what was the initial pH of the HCl solution? g
Chemistry
1 answer:
Mnenie [13.5K]3 years ago
3 0

Answer:

The Ph is 3

Explanation:

First of all, calculate the concentration of the acid.

The balanced equation of reaction is:

HCl + NaOH ===> NaCl + H2O

Using titration equation of formula

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 1 (from the balanced equation of reaction)

NB is the number of mole of base = 1 (from the balanced equation of reaction)

CA is the concentration of acid = ?

CB is the concentration of base = 0.1M

VA is the volume of acid = 100 mL

VB is the volume of base = 1mL

Substituting

CA×100/0.1×1 = 1/1

Therefore CA =0.1×1×1/100×1

CA = 0.001 M (this is the hydrogen ion concentration)

Ph of the acid is calculated using

Ph = -log [H+]

= -log 0.001

= 3

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Answer:

0.175\; \rm mol \cdot L^{-1}.

Explanation:

Magnesium chloride and silver nitrate reacts at a 2:1 ratio:

\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s).

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to  \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}.

The precipitate silver chloride \rm AgCl is insoluble in water and barely ionizes. Hence, \rm AgCl\! isn't rewritten as ions.

Net ionic equation:

\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}.

Calculate the initial quantity of nitrate ions in the mixture.

\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}.

Since nitrate ions \rm {NO_3}^{-} do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol.

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be (1/2) of the concentration in the original solution.

\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}.

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If 42.7 of 0.208 M hydrochloric acid are needed to completely neutralize a solution of calcium hydroxide, how many grams of calc
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Answer:

0.329 g

Explanation:

In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.

In a neutralization reaction, water is produced as a product, as well as a salt that we obtain after we exchange the cations: calcium bonds to chloride and hydrogen bonds to hydroxide (the latter is the formation of water). This means we also produce calcium chloride as a product. The overall reaction represents this as:

Ca(OH)_2(aq)+2 HCl (aq)\rightarrow CaCl_2 (aq)+2 H_2O (l)

Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:

c_{HCl}=\frac{n_{HCl}}{V_{HCl}}

Rearranging for moles of HCl, n:

n_{HCl}=c_{HCl}V_{HCl}

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Now that we have the expression for moles, we may also express moles of calcium hydroxide as the ratio between its mass and molar mass:

n_{Ca(OH)_2}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}

Using the last two equations, we see that:

\frac{1}{2}c_{HCl}V_{HCl}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}\\\therefore m_{Ca(OH)_2}=\frac{1}{2}c_{HCl}V_{HCl}M_{Ca(OH)_2}

Substitute the given data, as well as the molar mass of calcium hydroxide:

m_{Ca(OH)_2}=\frac{1}{2}\cdot0.208 M\cdot0.0427 L\cdot74.093 g/mol=0.329 g

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