Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 75

What volume of \(0.0200 M\) calcium hydroxide is required to neutralize \(35.00 \mathrm{~mL}\) of \(0.0500 \mathrm{M}\) nitric acid?

Short Answer

Expert verified
The volume of \(0.0200 M\) calcium hydroxide required to neutralize \(35.00 mL\) of \(0.0500 M\) nitric acid is \(43.75 mL\).

Step by step solution

01

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between calcium hydroxide (Ca(OH)2) and nitric acid (HNO3) is: \(Ca(OH)_{2_(aq)} + 2 HNO_{3_(aq)} \rightarrow Ca(NO_{3})_{2_(aq)} + 2 H_{2}O_{(l)}\) This tells us that one mole of calcium hydroxide reacts with two moles of nitric acid to produce one mole of calcium nitrate and two moles of water.
02

Calculate the moles of nitric acid

Given information: Volume of nitric acid = 35.00 mL Molarity of nitric acid = 0.0500 M To find the moles of nitric acid, use the formula: moles = molarity × volume (in liters) First, convert the volume of nitric acid from mL to L: \(35.00\mathrm{~mL} = 35.00 \times 10^{-3} L = 0.03500 L\) Now, calculate the moles of nitric acid: moles of nitric acid = 0.0500 M × 0.03500 L = 0.00175 moles
03

Find the moles of calcium hydroxide required for neutralization

From the balanced chemical equation, we see that the mole ratio of calcium hydroxide to nitric acid is 1:2. Moles of calcium hydroxide = (moles of nitric acid) / 2 Moles of calcium hydroxide = 0.00175 moles / 2 = 0.000875 moles
04

Calculate the volume of calcium hydroxide required

Molarity of calcium hydroxide = 0.0200 M Moles of calcium hydroxide = 0.000875 moles To find the volume of calcium hydroxide, use the formula: volume (in L) = moles / molarity Volume of calcium hydroxide (in L) = 0.000875 moles / 0.0200 M = 0.04375 L Finally, convert the volume of calcium hydroxide from L to mL: Volume of calcium hydroxide = 0.04375 L × 10^3 mL/L = 43.75 mL
05

Conclusion

It requires 43.75 mL of 0.0200 M calcium hydroxide to neutralize 35.00 mL of 0.0500 M nitric acid.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(10 . \mathrm{g}\) of \(\mathrm{AgNO}_{3}\) is available, what volume of \(0.25 \mathrm{MAgNO}_{3}\) solution can be prepared?

A \(500.0-\mathrm{mL}\) sample of \(0.200 M\) sodium phosphate is mixed with \(400.0 \mathrm{~mL}\) of \(0.289 M\) barium chloride. What is the mass of the solid produced?

What acid and what base would react in aqueous solution so that the following salts appear as products in the formula equation? Write the balanced formula equation for each reaction. a. potassium perchlorate b. cesium nitrate c. calcium iodide

A \(230 .-\mathrm{mL}\) sample of a \(0.275-M \mathrm{CaCl}_{2}\) solution is left on a hot plate overnight; the following morning, the solution is \(1.10 M\). What volume of water evaporated from the \(0.275 \mathrm{M} \mathrm{CaCl}_{2}\) solution?

A \(10.00-\mathrm{g}\) sample consisting of a mixture of sodium chloride and potassium sulfate is dissolved in water. This aqueous mixture then reacts with excess aqueous lead(II) nitrate to form \(21.75 \mathrm{~g}\) of solid. Determine the mass percent of sodium chloride in the original mixture.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free