`{:("List-I","List-II"),((A) 1xx10^(-8) "M KOH",P.pH=1.3),((B)1xx10^(-8)"MHBr" ,Q.pH=6.95),((C )1"MHCI"+1 "MNaOH",R.pH=7.0414),((D) 0.02 MH_(2)SO_(4),S.pH=7),(,"acidic solution"):}`

Answer

To solve the problem of matching the solutions with their respective pH values, we will analyze each solution in List-I and calculate their pH values step by step.

### Step 1: Analyze 1 x 10^(-8) M KOH
- KOH is a strong base and dissociates completely in water.
- The concentration of OH⁻ ions from KOH is 1 x 10^(-8) M.
- However, we must also consider the contribution of OH⁻ ions from the autoionization of water, which is 1 x 10^(-7) M.
- Total OH⁻ concentration = 1 x 10^(-8) + 1 x 10^(-7) = 1.1 x 10^(-7) M.
- Calculate pOH:
\[ \text{pOH} = -\log(1.1 \times 10^{-7}) \approx 6.96 \]
- Calculate pH:
\[ \text{pH} = 14 - \text{pOH} \approx 14 - 6.96 = 7.04 \]
- Match: A → R (pH = 7.04)

### Step 2: Analyze 1 x 10^(-8) M HBr
- HBr is a strong acid and dissociates completely in water.
- The concentration of H⁺ ions from HBr is 1 x 10^(-8) M.
- We must also consider the contribution of H⁺ ions from water, which is 1 x 10^(-7) M.
- Total H⁺ concentration = 1 x 10^(-8) + 1 x 10^(-7) = 1.1 x 10^(-7) M.
- Calculate pH:
\[ \text{pH} = -\log(1.1 \times 10^{-7}) \approx 6.95 \]
- Match: B → Q (pH = 6.95)

### Step 3: Analyze 1 M HCl + 1 M NaOH
- HCl is a strong acid and NaOH is a strong base. They will neutralize each other.
- The reaction is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- Since both are in equal concentrations, the resulting solution is neutral.
- Therefore, pH = 7.
- Match: C → S (pH = 7)

### Step 4: Analyze 0.02 M H₂SO₄
- H₂SO₄ is a strong acid and dissociates completely.
- The first dissociation gives 2 H⁺ ions for every mole of H₂SO₄.
- Therefore, the concentration of H⁺ ions = 0.02 M x 2 = 0.04 M.
- Calculate pH:
\[ \text{pH} = -\log(0.04) \approx 1.4 \]
- However, we also need to consider the contribution from water, which is negligible in this case.
- Match: D → P (pH = 1.3)

### Final Matching
- A → R (pH = 7.04)
- B → Q (pH = 6.95)
- C → S (pH = 7)
- D → P (pH = 1.3)