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Chapter 10: Q. 10.63 (page 355)

At rest, the $[H_{3} O^{+}]$ of the stomach fluid is $2.0 脳 10^{- 4} M$ . What is the $pH$ of the stomach fluid?

Short Answer

Expert verified

The pH text of the stomach fluid is 3.70.

Step by step solution

01

Step - 1 Introduction

The solution's $pH$ can be defined as the negative logarithmic (base 10) of the $[H_{3} O^{+}]$ concentration .

$pH = - \log [H_{3} O^{+}]$

$pOH = - \log [{OH}^{-}]$

$pH + pOH = 14$

02

Step - 2 Explanation

The solution has an acid concentration of $2.0 脳 10^{- 4} M$ . Calculate the solution's $pH$ using the following formula:

p H = - \log [H_{3} O^{*}]

$= - \log [2.0 脳 10^{- 4}]$

$= 4 - \log (2.0)$

$= 3.70$

As a result, the pH text of the stomach fluid is 3.70.

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Most popular questions from this chapter

Indicate whether each of the following solutions is acidic, basic, or neutral:

a. $[H_{3} O^{+}] = 2.0 脳 10^{- 5} M$

b. $[H_{3} O^{+}] = 1.4 脳 10^{- 9} M$

c. $[{OH}^{-}] = 8.0 脳 10^{- 3} M$

d. $[{OH}^{-}] = 3.5 脳 10^{- 10} M$

If you know the $[{OH}^{-}]$ , how can you determine the pH of a solution?

State whether each of the following is acidic, basic, or neutral:

a. soda $p H 13.22$

b. shampoo, $p H 5.7$

c. laundry detergent, $p H 9.44$

d. rain $p H 5.8$

e. honey, $p H 3.9$

f. cheese, $pH 4.9$

Balance each of the following neutralization equations:

a. ${HNO}_{3} (a q) + Ba (OH)_{2} (s) 鉄� H_{2} O (l) + Ba {({NO}_{3})}_{2} (a q)$

b.localid="1654005825158" $H_{2} {SO}_{4} (a q) + Al (OH)_{3} (s) 鉄� H_{2} O (l) + {Al}_{2} {({SO}_{4})}_{3} (a q)$

Calculate the volume, in millilitres, of a 0.215 M KOH solution that will completely neutralize each of the following:

a. 2.50 mL of a $0.825 M H_{2} {SO}_{4}$ solution

b. 18.5 mL of a $0.560 M {HNO}_{3}$ solution

c. 5.00 mL of a $3.18 M HCl$ solution

See all solutions

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