Answered: Calculate the pH during the titration…

Calculate the pH during the titration of 43.0 mL 0.108 M nitrous acid (pK, = 3.15) with 0.104 M NAOH after the addition of 11.0mL of the titrant. %3D 2.66 2.40 2.83 2.31 2.75

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter7: Neutralization Titrations And Graphical Representations

Section: Chapter Questions

Problem 8P

See similar textbooks

Related questions

Q: A 2.70 g sample of benzoic acid was dissolved in water to give 72.3 mL of solution. This solution…

Q: 3.) What is the pH at the equivalence point for the titration of 50.0 mL 0.20 M nitrous acid by 0.20…

Q: Consider the titration of 100.0 mL of 0.100 M H2NNH2 (Kb = 3.0 ✕ 10-6) by 0.200 M HNO3. Calculate…

A: We are authorized to answer three subparts at a time, since you have not mentioned which part you…

Q: A 40.00 mL sample of 0.1000 M diprotic malonic acid is titrated with 0.0900 M KOH. What volume KOH…

Q: 6. Calculate pl for the titration of 100.0 mL of 0.1023 M I´ with 0.1190 M AgNO3 for the addition of…

A: Reaction which will take place will be as follows: I- + AgNO3 -----> AgI + NO3- Once we start…

Q: In titrating 10 ml of 0.1 N HA (pKa-4),the pH after the addition of 12 ml 0.1N NaOH is O 3.69 О 11.2…

A: Given Normality of HA = 0.1 N Volume of HA = 10 mL Mole of HA = Normality of HA * Volume of HA Mole…

Q: What is the pH at the equivalence point of the titration of 60.0 mL of 0.0100 M HN3 with 0.0400 M…

A: HN3 + LiOH ------> LiN3 + H2O Number of mmols of HN3 = Number of mmols LiOH Number of mmols =…

Q: Consider the titration of 100.0 mL of 0.100 M H2NNH2 (Kb = 3.0 ✕ 10-6) by 0.200 M HNO3. Calculate…

A: Since there are multiple sub-parts, we will answer for only first three sub-parts. If you require…

Q: 3. 40.00 mL of 0.0900 M NAOH is diluted to 100mL and titrated with 0.1000M HCI. Calculate the pH…

A: Since you have posted a question with multiple sub-parts, we will solve the first three subparts for…

Q: What is the pCl of the solution during the titration of 50.0 mL of 0.100 M NaCl with 0.100 M AgNO3…

A: The number of millimoles of sodium chloride is calculated as,

Q: A solution containing 100.00 mL of 0.1800 M acetic acid (Ka = 1.78 x 10-5) is being titrated with…

A: To Calculate: a) pH at the equivalence point b) pH After the addition of 55.00 mL of NaOH

Q: Calculate the pH of a 50.00 mL of 0.2000 M ammonia (Ka = 5.7 x 1010) solution after addition of…

A: We have to calculate the pH of given solution.

Q: Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH)3N (K, = 5.2 x 104),…

A: Since there are multiple questions, we will answer only the first one. If you require the answers to…

Q: . Calculate the pH values and draw the titration curve for the titration of 500 ml of 0.010 M acetic…

Q: Weak acid HMa (0.150 M) has 65.00 mL is titrated with 0.15 M NaOH. The Kaminoan scientist adds…

Q: For the titration: 30.0mL of 0.150 M NH3 (Kb = 1.8 x 10-5) with 0.250 M HBr Calculate the pH after…

A: For the given titration we have to calculate the pH

Q: -Calculate pCl for the titration of 100.0mL of 0.1000 M CI¯ with 0.1000 M AGNO3 for the addition of…

A: At 0.0mL addition of AgNO3 pCl = -log(0.1000) = 1.000

Q: A 0.448 g sample of propanoic acid (CH3CH2COOH, Ka = 1.3. x 10-5) was dissolved to a final volume of…

Q: 15.83 Consider the titration of 25.0 mL of 0.0200 M H,CO3 with 0.0250 M KOH. Calculate the pH after…

A: “Since you have posted a question with multiple sub-parts, we will solve first three subparts for…

Q: 30.00 mL of 0.125 M ammonia (Kb = 1.8 x 10-5) is titrated with 0.0750 M HCl. Calculate the…

Q: 85.0 mL of 0.20 M HF (Ka = 7.2 x 10-4)is titrated with 5.0 mL of 0.65 M NaOH . Calculate the pH at…

Q: 4. Consider titration of 0.50L of 0.10 M acid HA (Ka = 5.1 x 10-8) with %3D 0.50 M KOH. a) What is…

A: The question involves the concepts of titration. we are titrating a weak acid with a strong base.…

Q: Calculate the pH during the titration of 50.00 mL of 0.0500 M NaOH with 0.100 M HCl at 25oC after…

Q: For the titration of 50.00 mL of 0.106 M NHg with 0.225 M HI, calculate the pH when neutralization…

A: Moles of NH3 = (vol in L*molarity in M) = (50*10^-3*0.106) mol = 0.0053 mol Half neutralization =…

Q: 0.0 mL sample of 0.100 M ammonium chloride (NH₂, K, 5.70 x 10 10) is titrated with 0.100 M NaOH.…

Q: In the titration of 20 mL CH3COOH (Ka = 1.76 x 10-5) 0.4 M with a solution of NaOH 0.4 M, calculate…

A: The buffer is the solution of reagent that is used to maintain its pH (potential of hydrogen) when…

Q: Calculate pcl for the titration of 100.0 ml of 0.1000 M cl with 0.1000M AgNo3 for the addition of…

Q: Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HC1. Calculate the pH after the…

A: Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for…

Q: Calculate the pH in the titration of 50.00 mL of 0.060 M acetic acid (CH3COOH) with a 0.120M sodium…

A: (A)pH when 0mL of NaOH solution is added:

Q: Consider the titration of 40.0 mL of 0.0600 M HONH2 (a weak base; Kb = 1.10e-08) with 0.100 M HI.…

Q: OOOO A 100.0-mL aliquot of 0.100 M diprotic acid H2A (pK1 =4.00, pK2 = 8.00) was titrated with 1.00…

A: The given acid is diprotic acid but at the first equivalent point, an equal mole of the acid reacts…

Q: In the titration of 50.0 ml of 0.1 M formic acid (ka- 1.8x 10) with 0.1M NAOH, what will be the pH…

Q: Calculate the pH during the titration of 50.00 mL of 0.0500 M NaOH with 0.1000 M HCl after the…

A: 50 mL of 0.0500 M of NaOH titrated against 0.100 M HCl at different volumes.

Q: 25.00 mL of 0.1000 M NH3 is titrated with 0.1000 M HCI. The K₁ of NH3 is 1.75 × 10¹5. Calculate pH…

Q: Consider the titration of 40.0 mL of 0.1000 M diprotic acid H2A (Ka1 = 2.93 × 10-5, Ka2 = 3.49 ×…

A: In the case of a weak acid versus a strong base, the pH is not neutral at the equivalence point.

Q: Calculate the silver lon concentration in terms of pAg during the titration of 50.00 ml. of 0.05000…

A: Given in following question 50ml of 0.05000M Nacl with 0.1000M AgNO3 after the addition calculate…

Q: 50.0 mL of 0.31 M acrylic acid (C2H3COOH, Ka = 1.71 x 10 ) solution is titrated with 0.31 M NAOH to…

Q: 18.00 mL 0.321 M what is the pH value of the solution when titration of NH3 a) 10.00 mL and b)…

A: The question is based on the concept of volumetric analysis. It involves titration of weak base…

Q: Derive a titration curve for the titration of 50 mL of 0.1 M NaCl with 0.1 M AGNO3, KSP (AGCI) =…

A: a) before adding AgNO3 : NaCl → Na++ Cl- 0.1 0.1 0.1 pCl = - log [ Cl-] = - log 0.1 =…

Q: A 40.00 mL sample of 0.1000 M diprotic malonic acid is titrated with 0.0900 M KOH. What volume KOH…

Q: Falculate the pH at the equivalence point for the titration of 0.20 M HCI with 0.20 M NH 3 (Kb = 1.8…

A: To calculate the pH at the equivalence point when 0.20 M HCl is titrated with 0.20 M NH3 Kb = 1.8 ×…

Q: 16.0mL of a 0.750M solution of H2CO3 are titrated with a 1.00M solution of KOH. At what volume of…

Q: The pH at 10, 100, and 110% titration (% of the equivalence point volume) for the titration of…

A: Solutions Given that the pH at 10, 100, 110% titration.

Q: 4. In the titration of15.00 mL of 0.200 M HC with 0.100 M NaOH, calculate the pH after the addition…

Q: 10.00 mL of 0.2099 M iodic acid {Ka = 1.70 xx 10^(-1)} is titrated with 0.1057 M NaOH. a) What is…

A: The pH of the solution is the power of protons in the solution. While pOH is related to the basicity…

Q: A titration is carried out for 25.0 mL of0.300M HA (Ka = 3.5 x 10-6) with 0.300M NAOH. Calculate the…

Q: pH 1X10 7. Consider the titration curve for the titration of 15 ml of Na,CO, as Na,A with 0.T 12 loa…

A: When a dibasic compound is titrated against a monoprotic acid then two peaks give the value of two…

Q: Find the pH and the volume (mL) of 0.527 M HNO3 needed to reach the equivalence point in the…

Q: Calculate the pH during the titration of 20.00 mL of 0.1 M NaOH with 0.100 M HCl after the addition…

A: Given - Volume of NaOH = 20.00 ml Molarity of NaOH= 0.1 M Molarity of HCl = 0.100 M

Concept explainers

Precipitation Titration

Precipitation analysis is used to analyze drugs, and to measure the salt content in food and beverages. It is an important concept from the examination point of view many practical questions are asked on the topic. This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

Question

Calculate the pH during the titration of 43.0 mL 0.108 M nitrous acid (pK, = 3.15) with 0.104 M NAOH after the addition of 11.0mL of the
titrant.
%3D
2.66
2.40
2.83
2.31
2.75

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 105) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. a. 0.0 mL b. 50.0 mL c. 100.0 mL d. 150.0 mL e. 200.0 mL f. 250.0 mL
Calculate the volume of 0.225-M NaOH required to titrate to the equivalence point for each of these samples. (a) 20.0 mL of 0.315-M HBr (b) 30.0 mL of 0.250-M HClO4 (c) 6.00 g of concentrated acetic acid, CH3COOH, which is 99.7% pure
Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a) before any NaOH is added? (b) at half-neutralization? (c) at the equivalence point? (d) when 0.10 mL less than the volume of NaOH to reach the equivalence point is added? (e) when 0.10 mL more than the volume of NaOH to reach the equivalence point is added? (f) Use your data to construct a plot similar to that shown in Figure 14.10 (pH versus volume NaOH added).
Without doing detailed calculations, sketch the curve for the titration of 30.0 mL of 0.10 M NaOH with 0.10 M HCl. Indicate the approximate pH at the beginning of the titration and at the equivalence point. What is the total solution volume at the equivalence point?
Consider the titration of 300.0 mL of 0.450 M NH3 (Kb = 1.8  10–5) with 0.450 M HNO3. At the stoichiometric point of this titration, the pH is:
If 0.3074 g of a mixture of pure KCl and KBr requires 30.98 mL of 0.1007 M AgNO3 solution for its titration, calculate the% KCl and KBr in the sample. (Pat Data: Cl, 35.453; Br, 79.904; K, 39.098)
Argentometric titration of a 0.7908-g sample containing chloride ion required 45.32 mL of 0.1046 M AgNO3. Express the results of this analysis in terms of the percentage ofa) Cl–b) BaCl2.H2O. c) ZnCl2.2NH4Cl
Calculate the pH during the titration of 40.00 mL of 0.0250 M KOH with 0.0500 M HCl after addition of the following volumes of titrant reagent (HCl): a) 0.00 mL b) 10.00 mL c) 15.00 mL d) 18.00 mL e) 20.00 mL f) 25.00 mL
In the quadruplicate titration of 0.2068 g; 0.1997g; 0.2245 g and 0.2137 g of sodium carbonate 36.31 mL were spent; 35.11mL; 39.00 mL and 37.54 mL of perchloric acid. a) Calculate the average molarity of the perchloric acid solution. b) Would there be statistical justification for rejecting the furthest molarity? Tabular Q = 0.829 for n = 4 and 95% confidence.
The experimental Ksp of calcium hydroxide is _________ x 10-6 if 23.09mL of the sat. calcium hydroxide solution was titrated with 12.53mL of 0.049M HCl(aq), considering 4 as an exact value for the calculation?
Titration of 50.00 mL of 0.04715 M Na2C2O4 required 39.25 mL of a potassium permanganate solution. 2MnO4- + 5C2O4-2 +16H+ -> 2Mn2+ + 10CO2(g) + 8H2O Calculate the molar concentration of the KMnO4 solution. Pls show step by step. What type of molecules will interfere in this analysis ?
Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid: (a) 0 mL (b) 18.00 mL (c) 30.00 mL (d) 35.10 mL