Question

calculate the ph of each aqueous solution. round each of your answer to 2 decimal places.
note: the ion - product constant for water, ( k_{w} ), is ( 1.0\times10^{-14} ) at ( 25^{circ}c ).
part 1 of 2
0.000018 m hcl.
( ph=)
part 2 of 2
0.030 m koh.
( ph=)

Explanation:

Part 1 of 2: For \(0.000018\ M\ HCl\)

Step1: Determine \([H^{+}]\)

Since \(HCl\) is a strong acid, it dissociates completely: \(HCl
ightarrow H^{+}+Cl^{-}\). So, \([H^{+}]=0.000018\ M = 1.8\times10^{- 5}\ M\)

Step2: Calculate \(pH\)

Use the formula \(pH=-\log[H^{+}]\).

$$pH =-\log(1.8\times 10^{-5})$$

$$pH=-( \log(1.8)+\log(10^{-5}))$$

$$pH=-(0.255 - 5)$$

$$pH = 4.745\approx4.75$$

Part 2 of 2: For \(0.030\ M\ KOH\)

Step1: Determine \([OH^{-}]\)

Since \(KOH\) is a strong base, it dissociates completely: \(KOH
ightarrow K^{+}+OH^{-}\). So, \([OH^{-}]=0.030\ M\)

Step2: Calculate \([H^{+}]\)

Use the ion - product constant \(K_{w}=[H^{+}][OH^{-}]\), where \(K_{w}=1.0\times10^{-14}\). Then \([H^{+}]=\frac{K_{w}}{[OH^{-}]}\)

$$ [H^{+}]=\frac{1.0\times 10^{-14}}{0.030}=\frac{1.0\times 10^{-14}}{3.0\times 10^{-2}}=\frac{1}{3}\times10^{-12}\ M$$

Step3: Calculate \(pH\)

Use the formula \(pH =-\log[H^{+}]\)

$$pH=-\log(\frac{1}{3}\times10^{-12})$$

$$pH=-( \log(1)-\log(3)+\log(10^{-12}))$$

$$pH=-(0 - 0.477-12)$$

$$pH=12.477\approx12.48$$

Answer:

• Part 1 of 2: \(pH = 4.75\)
• Part 2 of 2: \(pH=12.48\)