6. 50 ml 0.1 M K_(2)CrO_(4) dititrasi dengan 0,2 M AgNO_(3) . Hitung harga pCrO_(4) bila volume titran berikut telah ditambahkan: a. 0 ml ; 25 ml; 49,9 ml; 50 ml ; 50,1 ml; 60 ml

## Titration of K2CrO4 with AgNO3: Calculating pCrO4This problem involves the titration of potassium chromate (K2CrO4) with silver nitrate (AgNO3). The reaction forms a precipitate of silver chromate (Ag2CrO4):**2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq)**To calculate the pCrO4 at different volumes of titrant added, we need to consider the following:1. **Initial concentration of CrO4^2-:** The initial concentration of chromate ions (CrO4^2-) is 0.1 M.2. **Stoichiometry of the reaction:** The reaction consumes 1 mole of K2CrO4 for every 2 moles of AgNO3.3. **Equilibrium constant:** The solubility product constant (Ksp) for Ag2CrO4 is 1.1 x 10^-12.**Calculations:****a. 0 ml AgNO3 added:*** No reaction has occurred yet.* [CrO4^2-] = 0.1 M* pCrO4 = -log(0.1) = 1**b. 25 ml AgNO3 added:*** Moles of AgNO3 added: 0.025 L * 0.2 mol/L = 0.005 mol* Moles of K2CrO4 reacted: 0.005 mol / 2 = 0.0025 mol* Moles of K2CrO4 remaining: 0.05 L * 0.1 mol/L - 0.0025 mol = 0.0025 mol* [CrO4^2-] = 0.0025 mol / 0.075 L = 0.0333 M* pCrO4 = -log(0.0333) = 1.48**c. 49.9 ml AgNO3 added:*** Moles of AgNO3 added: 0.0499 L * 0.2 mol/L = 0.00998 mol* Moles of K2CrO4 reacted: 0.00998 mol / 2 = 0.00499 mol* Moles of K2CrO4 remaining: 0.05 L * 0.1 mol/L - 0.00499 mol = 0.00001 mol* [CrO4^2-] = 0.00001 mol / 0.0999 L = 1 x 10^-4 M* pCrO4 = -log(1 x 10^-4) = 4**d. 50 ml AgNO3 added:*** This is the equivalence point. All K2CrO4 has reacted.* We need to use the Ksp to calculate [CrO4^2-]: * Ksp = [Ag+]^2 [CrO4^2-] = 1.1 x 10^-12 * [Ag+] = 2 * [CrO4^2-] (from the stoichiometry) * 1.1 x 10^-12 = (2[CrO4^2-])^2 [CrO4^2-] * [CrO4^2-] = (1.1 x 10^-12 / 4)^1/3 = 6.5 x 10^-5 M* pCrO4 = -log(6.5 x 10^-5) = 4.19**e. 50.1 ml AgNO3 added:*** Excess AgNO3 is present.* We need to calculate the concentration of Ag+ and use the Ksp to find [CrO4^2-]: * Moles of AgNO3 added: 0.0501 L * 0.2 mol/L = 0.01002 mol * Moles of AgNO3 in excess: 0.01002 mol - 0.01 mol = 0.00002 mol * [Ag+] = 0.00002 mol / 0.1001 L = 2 x 10^-4 M * Ksp = [Ag+]^2 [CrO4^2-] = 1.1 x 10^-12 * [CrO4^2-] = 1.1 x 10^-12 / (2 x 10^-4)^2 = 2.75 x 10^-5 M* pCrO4 = -log(2.75 x 10^-5) = 4.56**f. 60 ml AgNO3 added:*** Excess AgNO3 is present.* We can calculate [CrO4^2-] using the same method as in (e): * Moles of AgNO3 in excess: 0.06 L * 0.2 mol/L - 0.01 mol = 0.01 mol * [Ag+] = 0.01 mol / 0.11 L = 0.091 M * Ksp = [Ag+]^2 [CrO4^2-] = 1.1 x 10^-12 * [CrO4^2-] = 1.1 x 10^-12 / (0.091)^2 = 1.33 x 10^-10 M* pCrO4 = -log(1.33 x 10^-10) = 9.88**Summary:**| Volume AgNO3 (ml) | pCrO4 ||---|---|| 0 | 1 || 25 | 1.48 || 49.9 | 4 || 50 | 4.19 || 50.1 | 4.56 || 60 | 9.88 |This table shows how the pCrO4 changes during the titration. As more AgNO3 is added, the concentration of CrO4^2- decreases, leading to an increase in pCrO4. The equivalence point is reached when all K2CrO4 has reacted, and the pCrO4 is determined by the Ksp of Ag2CrO4. After the equivalence point, the excess AgNO3 further reduces the concentration of CrO4^2-, resulting in a significant increase in pCrO4.