Q 7 please 1. The pH of a solution is 4.80. What is the concentration of hydroxide ions in this solution? a.4.2 x 10 M b. 1.6 x 105 M c. 3.6 x 10-12 M d. 6.3 x 10-10 M 2 the pH of a 0.50 M solution of NaNOâ‚‚. Ka for HNOâ‚‚ is 4×10-5 a 12.1 b. 5.48 c. 1.82 c. 8.90 d 8.52 3- What was the pH of the solution that result from titration of 25.0 ml of 0.5 M solution of weak base (Kb – 9.74×10) wi 30 ml of 0.1 M hydrochloric acid, HCI.? a. 5.10 b. 4.92 d. 9.1 4-Which of the following combinations cannot produce a buffer solution? a. HNO2 and NaNOâ‚‚ b. HCN and NaCN c. HCIO4 and NaCIO4 d. NH3 and (NH4)2SO4 e. 2.0 x 10-8 M d. 5.5 e. NH3 and NH4Br 5- What is the pH at the equivalence point in the titration of 100.0 mL of 0.20 M ammonia (NH3) with 0.10 M hydrochloric acid (HCI)? Kb for NH31.8x 10-5 a. 4.6 b. 5.2 c. 7.0 e. 4.9 e.7.00 9. what will happened to the pH of buffer system when it diluted a. ApH=0 b. pH will increased c. pH will decrease 6- If 10.00 ml of 0.1M NaOH solution is needed to reach the end point of 10 ml HNO3. Then the concentration of HNO3 in ppm is: [M.wt. of HNO3= 63 g/mol] e. 25.77 x 10³ ppm a. 9.45×10³ ppm b.14.70 x10³ ppm c. 6.30 × 10³ ppm d. 18.77 × 10¹ ppm 7. A 0.6745-gram sample of KHP reacts with 41.75 mL of KOH solution for complete neutralization. What is the molarity of the KOH solution? (Molecular weight of KHP = 204 g/mol. KHP has one acidic hydrogen.) a. 0.158 M c. 0.139 M b. 0.099 M d. 0.079 M e. 0.061 M 8. Which of the following combinations would give a pH =7.00 at the “equivalence point” (when equal moles of each have been added)? a. HCI + KF b. HCN + NaOH c. HF + HCI d. HCI + KOH 13. The molar analytical concentration of NO were added to 20.0 ml of 0. 82 M Fe (NO3)2 a.1.09M b. 2.53 M d. pH >7 e. pH = 7 10. the pH for Acidic buffer system with highest capacity if the Ka for the weak Acid 1.8 104 is a. 4.3 b.5.3 c.3.77 d.6.8 e. 8.3 e. 9.5 c.0.73 M 11.The ppm concentration of Cl-ion in 100 ml mixture solution of 0.01 M CaCl2 and 0.1 M HCI is (MW of Cl = 35.45 g/mol and for Cl =35.45 g/mol ) is :- b. 4254 ppm c. 355 ppm d. 9150ppm a. 243 ppm e. 850ppm from the primary- 12. The mass of AgNO3(s) needed to prepare 1.000 L of 0.0500 M AgNO3 (169.87 g/mol) solution standard-grade solid is a. 18.548g b. 16.987 g c. 16.139 g d. 9.843g e. 10.36 g ions in the solution produced when 25.0 ml of Distilled water d. 0.11 M 14. Standard solution of EDTA (0.100 M) is being used to titrate 25.00 ml of (0.0100) M of Zn²+ solution (buffered at pH =10, 04-0.36 and kMY = 3.2×10¹6) the pZn after the addition of 25 ml EDTA solution is (at equivalent point) a. 8.68 b. 9.03 d. 9.79 c. 3.76 e. 0.13 M e. 6.89
The Correct Answer and Explanation is :
Question 7:
A 0.6745-gram sample of KHP reacts with 41.75 mL of KOH solution for complete neutralization. The molarity of the KOH solution can be calculated using the molecular weight of KHP, which is 204 g/mol. KHP reacts with KOH in a 1:1 molar ratio since KHP contains one acidic hydrogen.
Step-by-step solution:
- Calculate moles of KHP:
[
\text{Moles of KHP} = \frac{\text{Mass of KHP}}{\text{Molar mass of KHP}} = \frac{0.6745 \, \text{g}}{204 \, \text{g/mol}} = 0.003307 \, \text{mol}
] - Determine the molarity of KOH:
Since the moles of KHP are equal to the moles of KOH due to the 1:1 stoichiometry:
[
\text{Molarity of KOH} = \frac{\text{Moles of KOH}}{\text{Volume of KOH in liters}}
]
Volume of KOH in liters:
[
41.75 \, \text{mL} = 0.04175 \, \text{L}
]
Substituting:
[
\text{Molarity of KOH} = \frac{0.003307}{0.04175} = 0.079 \, \text{M}
]
Final Answer:
d. 0.079 M
Explanation:
The calculation revolves around stoichiometry and molarity. KHP (Potassium Hydrogen Phthalate) is a primary standard acid used in titrations due to its high purity and one acidic proton. The reaction with KOH is straightforward:
[
\text{KHP} + \text{KOH} \to \text{KP} + \text{H}_2\text{O}
]
We first find the moles of KHP using its mass and molar mass, then calculate the molarity of KOH based on the volume of KOH used. The final molarity of 0.079 M fits perfectly with the neutralization reaction and the given data. This step-by-step method ensures accuracy, critical for quantitative chemical analysis.