Question

for the acids, use the formula ph = -logh⁺. (because hcl has 1 h⁺ ion per formula unit, h⁺ equals the molarity.) for water, use what you know about the ph of neutral solutions. for the bases, examine and extend the pattern. retry
0.1 m hcl
0.001 m hcl
0.00001 m hcl
distilled water
0.00001 m naoh
0.001 m naoh
0.1 m naoh
ph:
ph:
ph:
ph:
ph:
ph:
ph:

Explanation:

Step1: Calculate pH of 0.1 M HCl

Since $[H^+]=0.1\ M$, $pH = -\log[H^+]=-\log(0.1)= 1$.

Step2: Calculate pH of 0.001 M HCl

$[H^+]=0.001\ M$, $pH = -\log(0.001)=3$.

Step3: Calculate pH of 0.00001 M HCl

$[H^+]=0.00001\ M$, $pH = -\log(0.00001)=5$.

Step4: Determine pH of distilled water

Distilled water is neutral, so $pH = 7$.

Step5: Calculate pOH of 0.00001 M NaOH

For bases, first find pOH. Since $[OH^-]=0.00001\ M$, $pOH=-\log(0.00001) = 5$. Then $pH=14 - pOH=14 - 5 = 9$.

Step6: Calculate pOH of 0.001 M NaOH

$[OH^-]=0.001\ M$, $pOH=-\log(0.001)=3$. Then $pH=14 - 3=11$.

Step7: Calculate pOH of 0.1 M NaOH

$[OH^-]=0.1\ M$, $pOH=-\log(0.1)=1$. Then $pH=14 - 1 = 13$.

Answer:

0.1 M HCl: 1
0.001 M HCl: 3
0.00001 M HCl: 5
Distilled Water: 7
0.00001 M NaOH: 9
0.001 M NaOH: 11
0.1 M NaOH: 13