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charle [14.2K]
2 years ago
10

A titration is performed on a 25.0 mL sample of calcium hydroxide. A volume of

Chemistry
1 answer:
slamgirl [31]2 years ago
6 0

Answer:

\boxed {\boxed {\sf 0.28 \ M}}

Explanation:

Titration is a method used to determine the concentration of a substance. The  formula for this is:

M_AV_A= M_BV_B

Where M is the molarity of the acid or base and V is the volume of the acid or base.

We know that 46.0 milliliters of a 0.15 molar solution of nitric acid wereused in the titration. 25.0 milliliters of solution of calcium hydroxide of unknown molarity were also used.

\bullet \  M_A= 0.15 \ M \\\bullet \  V_A= 46.0 \ mL\\\bullet \  V_B= 25.0 \ mL

Substitute these known values into the formula.

0.15 \ M * 46.0 \ mL= M_B * 25.0 \ mL

We are solving for the molarity of the base, so we must isolate the variable M_B. It is being multiplied by 25.0 milliliters. The inverse operation of multiplication is division, so we divide both sides of the equation by 25.0 mL.

\frac {0.15 \ M * 46.0 \ mL}{25.0 \ mL}= \frac{M_B * 25.0 \ mL}{25.0 \ mL}

\frac {0.15 \ M * 46.0 \ mL}{25.0 \ mL}= M_B

The units of milliliters (mL) cancel.

\frac {0.15 \ M * 46.0 }{25.0}=M_B

\frac {6.9}{25.0 } \ M =M_B

0.276 \ M = M_B

The original measurements have 2 or 3 significant figures. We always round our answer to the least number of sig figs, which is 2. For the number we calculated, that is the hundredths place. The 6 in the thousandths place tells us to round the 7 up to an 8.

0.28 \ M \approx M_B

The concentration of calcium hydroxide is approximately <u>0.28 M.</u>

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Answer:

V=0.0310L=3.10mL

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

n_{acid}=n_{KOH}

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n_{acid}=1.70g*\frac{1mol}{84.48g}=0.0201mol

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V=\frac{0.0201mol}{0.650mol/L}\\\\V=0.0310L=3.10mL

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3 years ago
NaOH + HCl → NaCl + H2O
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0.00496 g H2O

Explanation:

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Which of the following is a double replacement reaction? Ca(OH)2 + H2SO4 → CaSO4 + 2H2O CH4 + 2O2 → CO2 + 2H2O 8Fe + S8 → 8FeS Z
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That would be the first option Ca(OH)2 + H2SO4  → CaSO4 + 2H2O.

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Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by pheno
melamori03 [73]

1.062 mol/kg.

<em>Step 1</em>. Write the balanced equation for the neutralization.

MM = 204.22 40.00

KHC8H4O4 + NaOH → KNaC8H4O4 + H2O

<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)

Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)

= 4.035 mmol KHP

<em>Step 3</em>. Calculate the moles of NaOH

Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)

= 4.035 mmol NaOH

<em>Step 4</em>. Calculate the mass of the NaOH

Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)

= 161 mg NaOH

<em>Step 5</em>. Calculate the mass of the water

Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g

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<em>Step 6</em>. Calculate the molal concentration of the NaOH

<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg

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