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General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

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BuyFindarrow_forward

General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

What is the pH of a solution in which 15 mL of 0.10 M NaOH is added to 25 mL of 0.10 M HCl?

Interpretation Introduction

Interpretation:

The pH of the solution prepared by addition of 15 mL of 0.10 M NaOH with 25 mL of 0.10 M HCl has to be calculated

Concept Introduction:

pH definition:

The pH of a solution is defined as the negative base-10 logarithm of the hydronium ion [H3O+] concentration.

pH=-log[H3O+]

To Calculate: The pH of the solution prepared by addition of 15 mL of 0.10 M NaOH with 25 mL of 0.10 M HCl

Explanation

Given data:

The concentration of NaOH = 0.10 M

The concentration of HCl = 0.10 M

The volume of NaOH = 15 mL

The volume of HCl = 25 mL

pH calculation:

All of the hydroxide ion OH reacts with the hydronium ion H3O+ from HCl .

The moles are calculated as follows,

mol H3O+ =(0.10 mol HCl/L) × 0.025 L HCl =0.0025 mol H3O+mol OH =(0.10 mol NaOH/L) × 0.015 L NaOH =0.0015 mol OH

The moles of hydronium ion remaining can be obtained by subtracting the initial moles of hydronium ion by the moles of hydroxide ion reacted.

mol H3O+ left = intial mol of H3O+ mol of OH- =0

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Chapter 16 Solutions

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Sect-16.4 P-16.9ESect-16.4 P-16.3CCSect-16.5 P-16.4CCSect-16.5 P-16.10ESect-16.5 P-16.11ESect-16.6 P-16.12ESect-16.6 P-16.13ESect-16.6 P-16.5CCSect-16.6 P-16.6CCSect-16.7 P-16.14ESect-16.7 P-16.15ESect-16.7 P-16.16ECh-16 P-16.1QPCh-16 P-16.2QPCh-16 P-16.3QPCh-16 P-16.4QPCh-16 P-16.5QPCh-16 P-16.6QPCh-16 P-16.7QPCh-16 P-16.8QPCh-16 P-16.9QPCh-16 P-16.10QPCh-16 P-16.11QPCh-16 P-16.12QPCh-16 P-16.13QPCh-16 P-16.14QPCh-16 P-16.15QPCh-16 P-16.16QPCh-16 P-16.17QPCh-16 P-16.18QPCh-16 P-16.19QPCh-16 P-16.20QPCh-16 P-16.21QPCh-16 P-16.22QPCh-16 P-16.23QPCh-16 P-16.24QPCh-16 P-16.25QPCh-16 P-16.26QPCh-16 P-16.27QPCh-16 P-16.28QPCh-16 P-16.29QPCh-16 P-16.30QPCh-16 P-16.31QPCh-16 P-16.32QPCh-16 P-16.33QPCh-16 P-16.34QPCh-16 P-16.35QPCh-16 P-16.36QPCh-16 P-16.37QPCh-16 P-16.38QPCh-16 P-16.39QPCh-16 P-16.40QPCh-16 P-16.41QPCh-16 P-16.42QPCh-16 P-16.43QPCh-16 P-16.44QPCh-16 P-16.45QPCh-16 P-16.46QPCh-16 P-16.47QPCh-16 P-16.48QPCh-16 P-16.49QPCh-16 P-16.50QPCh-16 P-16.51QPCh-16 P-16.52QPCh-16 P-16.53QPCh-16 P-16.54QPCh-16 P-16.55QPCh-16 P-16.56QPCh-16 P-16.57QPCh-16 P-16.58QPCh-16 P-16.59QPCh-16 P-16.60QPCh-16 P-16.61QPCh-16 P-16.62QPCh-16 P-16.63QPCh-16 P-16.64QPCh-16 P-16.65QPCh-16 P-16.66QPCh-16 P-16.67QPCh-16 P-16.68QPCh-16 P-16.69QPCh-16 P-16.70QPCh-16 P-16.71QPCh-16 P-16.72QPCh-16 P-16.73QPCh-16 P-16.74QPCh-16 P-16.75QPCh-16 P-16.76QPCh-16 P-16.77QPCh-16 P-16.78QPCh-16 P-16.79QPCh-16 P-16.80QPCh-16 P-16.81QPCh-16 P-16.82QPCh-16 P-16.83QPCh-16 P-16.84QPCh-16 P-16.85QPCh-16 P-16.86QPCh-16 P-16.87QPCh-16 P-16.88QPCh-16 P-16.89QPCh-16 P-16.90QPCh-16 P-16.91QPCh-16 P-16.92QPCh-16 P-16.93QPCh-16 P-16.94QPCh-16 P-16.95QPCh-16 P-16.96QPCh-16 P-16.97QPCh-16 P-16.98QPCh-16 P-16.99QPCh-16 P-16.100QPCh-16 P-16.101QPCh-16 P-16.102QPCh-16 P-16.103QPCh-16 P-16.104QPCh-16 P-16.105QPCh-16 P-16.106QPCh-16 P-16.107QPCh-16 P-16.108QPCh-16 P-16.109QPCh-16 P-16.110QPCh-16 P-16.111QPCh-16 P-16.112QPCh-16 P-16.113QPCh-16 P-16.114QPCh-16 P-16.115QPCh-16 P-16.116QPCh-16 P-16.117QPCh-16 P-16.118QPCh-16 P-16.119QPCh-16 P-16.120QPCh-16 P-16.121QPCh-16 P-16.122QPCh-16 P-16.123QPCh-16 P-16.124QPCh-16 P-16.125QPCh-16 P-16.126QPCh-16 P-16.127QPCh-16 P-16.128QPCh-16 P-16.129QPCh-16 P-16.130QPCh-16 P-16.131QPCh-16 P-16.132QPCh-16 P-16.133QPCh-16 P-16.134QPCh-16 P-16.135QPCh-16 P-16.136QPCh-16 P-16.137QPCh-16 P-16.138QPCh-16 P-16.139QPCh-16 P-16.140QPCh-16 P-16.141QPCh-16 P-16.142QPCh-16 P-16.143QPCh-16 P-16.144QPCh-16 P-16.145QPCh-16 P-16.146QPCh-16 P-16.147QPCh-16 P-16.148QPCh-16 P-16.149QPCh-16 P-16.150QPCh-16 P-16.151QPCh-16 P-16.152QPCh-16 P-16.153QPCh-16 P-16.154QPCh-16 P-16.155QPCh-16 P-16.156QPCh-16 P-16.157QP

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