Question

PLEASE HELP FAST!! WILL GIVE BRAINLIEST!!

Answer 1

0.1M HCl              [H⁺]=0.1=10⁻¹       pH= - log 10⁻¹ = 1,  pH = 1
0.001M HCl          [H⁺]=0.001=10⁻³       pH= - log 10⁻³ = 3,  pH = 3
0.00001M HCl      [H⁺]=0.00001=10⁻⁵       pH= - log 10⁻⁵ = 5,  pH = 5
Distilled water                                           pH=7


0.00001 NaOH   [OH⁻]=10⁻⁵, [OH⁻][H⁺]=10⁻¹⁴, [H⁺]=10⁻¹⁴/10⁻⁵=10⁻⁹,
                           pH=9

0.001 NaOH   [OH⁻]=10⁻³, [OH⁻][H⁺]=10⁻¹⁴, [H⁺]=10⁻¹⁴/10⁻³=10⁻¹¹,
                           pH=11

0.1 NaOH   [OH⁻]=10⁻¹, [OH⁻][H⁺]=10⁻¹⁴, [H⁺]=10⁻¹⁴/10⁻¹=10⁻¹³,
                           pH=13

Answer 2

Answer:

pH of 0.1M HCl=1

pH of 0.001M HCl=3

pH of 0.00001M HCl=5

pH of distilled water ,pH=7

pH of 0.00001 M NaOH =9

pH of 0.001 M NaOH=11

pH of 0.1 M NaOH=13.

Explanation:

We are given that For the acids, use the formula pH=-log[H+]

Because HCl has 1$H^+$ ion per formula unit

$[H+]$=Molarity

We are given that

0.1 M HCl means 0.1 molarity of H+ ions

[H+]=0.1M

Applying the given formula for calculating pH

Then pH of o.1MHCl=-log(0.1)

pH=-log$(1\times10^{-1})$

pH=-(log1-log10)

log 10=1, log 1=0

pH of 0.1M HCl=-(0-1)=1

pH of 0.001M HCl

pH=$-log(1\times 10^{-3})$

pH of 0.001M HCl=-(-3 log 10)=3

pH of 0.00001M HCl

pH=$-log(1\times 10^{-5})$

pH=-(-5)log10=5

pH of 0.00001M HCl=5

pH of distilled water ,pH=7 because water is neutral .

pH of 0.00001M NaOH

It means [OH-]=0.00001M=$1\times 10^{-5}$

We know that$[OH-][H+]=10^{-14}$

$[H+]=\frac{10^{-14}}{[OH-]}$

Substitute the value then we get the concentration of hydrogen ion

[H+]=$\frac{10^{-14}}{10^{-5}}$

[H+]=$10^{-14+5}$

Using identity:$\frac{a^x}{a^y}=a^{x-y}$

[H+]=$10^{-9}$

pH=-log(10^{-9})=-(-9) log 10=9

Hence, pH of 0.00001 M NaOH =9

pH of 0.001 M NaOH

$[OH-]=10^{-3}$

[H+]=\frac{10^{-14}}{10^{-3}}[/tex]

[H+]=$10^{-14+3}=10^{-11}$

pH=-log[H+]=-log(10^{-11})=11

Therefore, pH of 0.001 M NaOH=11

pHb of 0.1 M NaOH

[OH-]=$10^{-1}$

[H+]=$\frac{10^{-14}}{10^{-1}}$

[H+]=$10^{-14+1}=10^{-13}$

pH =$-(log 10^{-13})$

pH=-(-13 log 10)=13

Hence, pH of 0.1 M NaOH=13.

The pH of acids are less than the 7 and the pH value of bases is greater than the 7.