Calculate the volume in mL of a 1.420 M NaOH solution required to titrate the following solutions:25 mL of a 2.430 M HCl solution25 mL of a 4.500 M H2SO4 solution25 mL of a 1.500 M H3PO4 solution

(1). M1=2.430 M HCl,V1=25 mL=0.025 LM2=1.420 M NaOH, V2=?M1V1=M2V2V2=2.430×0.0251.420=0.04278…

Answered: Calculate the volume in mL of a 1.420 M…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Concept explainers

Question

Calculate the volume in mL of a 1.420 M NaOH solution required to titrate the following solutions:

25 mL of a 2.430 M HCl solution
25 mL of a 4.500 M H₂SO₄ solution
25 mL of a 1.500 M H₃PO₄ solution

Expert Solution

Step 1

$(1) . M_{1} = 2.430 M H C l, V_{1} = 25 m L = 0.025 L M_{2} = 1.420 M N a O H, V_{2} = ? M_{1} V_{1} = M_{2} V_{2} V_{2} = \frac{2.430 \times 0.025}{1.420} = 0.04278 L = 42.78 m L n e e d e d o f N a O H = 42.78 m L$

$(B) . w h e n M_{1} = 4.500 M H_{2} S O_{4} V_{1} = 25 m L = 0.025 L M_{2} = 1.420 M N a O H, V_{2} = ? \frac{M_{1} V_{1}}{1} = \frac{M_{2} V_{2}}{2} V_{2} = \frac{2 M_{1} V_{1}}{M_{2}} = \frac{2 \times 4.500 M \times 0.025 L}{1.420 M} = 0.1584 L V_{2} = 158.4 m L N e e d e d o f N a O H = 158.4 m L$

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions