Question

In: Chemistry

: Calculate the pH of a solution that results from 125 mL of 0.10M NH3 with...

: Calculate the pH of a solution that results from 125 mL of 0.10M NH3 with 250 mL of 0.10 M NH4Cl; Kb = 1.76 x 10-5

Calculate the pH when 0.010 mols of NaOH are added to 250.0 mL buffer containing 0.250 M acetic acid and 0.250 M sodium acetate. (pka = 4.74)

A 50.0 mL sample of 0.200M NaOH is titrated with 0.200M nitric acid. Calculate the pH after 25.0 mL HNO3 has been added.

Solutions

Expert Solution

1)

Concentration after mixing = mol of component / (total volume)

M(NH3) after mixing = M(NH3)*V(NH3)/(total volume)

M(NH3) after mixing = 0.1 M*125.0 mL/(125.0+250.0)mL

M(NH3) after mixing = 3.333*10^-2 M

Concentration after mixing = mol of component / (total volume)

M(NH4+) after mixing = M(NH4+)*V(NH4+)/(total volume)

M(NH4+) after mixing = 0.1 M*250.0 mL/(250.0+125.0)mL

M(NH4+) after mixing = 6.667*10^-2 M

Kb = 1.76*10^-5

pKb = - log (Kb)

= - log(1.76*10^-5)

= 4.754

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.754+ log {6.667*10^-2/3.333*10^-2}

= 5.056

use:

PH = 14 - pOH

= 14 - 5.0555

= 8.9445

Answer: 8.94

2)

mol of NaOH added = 0.01 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.25 M *0.25 L

mol of CH3COO- = 0.0625 mol

mol of CH3COOH = 0.25 M *0.25 L

mol of CH3COOH = 0.0625 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (0.0625 + 0.01) mol

mol of CH3COO- = 0.0725 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (0.0625 - 0.01) mol

mol of CH3COOH = 0.0525 mol

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.74+ log {7.25*10^-2/5.25*10^-2}

= 4.88

Answer: 4.88

3)

Given:

M(HNO3) = 0.2 M

V(HNO3) = 25 mL

M(NaOH) = 0.2 M

V(NaOH) = 50 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.2 M * 25 mL = 5 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2 M * 50 mL = 10 mmol

We have:

mol(HNO3) = 5 mmol

mol(NaOH) = 10 mmol

5 mmol of both will react

remaining mol of NaOH = 5 mmol

Total volume = 75 mL

[OH-]= mol of base remaining / volume

[OH-] = 5 mmol/75 mL

= 6.667*10^-2 M

use:

pOH = -log [OH-]

= -log (6.667*10^-2)

= 1.1761

use:

PH = 14 - pOH

= 14 - 1.1761

= 12.8239

Answer: 12.82


Related Solutions

78 A.) Calculate the pH of the solution that results from mixing 20.0 mL of 0.022...
78 A.) Calculate the pH of the solution that results from mixing 20.0 mL of 0.022 M HCN(aq) with 80.0 mL of 0.066 M NaCN(aq). pH= b. Calculate the pH of the solution that results from mixing 28.0 mL of 0.117 M HCN(aq) with 28.0 mL of 0.117 M NaCN(aq). C. Calculate the pH of the solution that results from mixing 66.0 mL of 0.057 MHCN(aq) with 34.0 mL of 0.034 M NaCN(aq).. The Ka value for HCN is 4.9×10^(−10)...
Calculate the pH of the solution that results from each of the following mixtures. 140.0 mL...
Calculate the pH of the solution that results from each of the following mixtures. 140.0 mL of 0.27 M  HF with 225.0 mL of 0.30 M  NaF 175.0 mL of 0.12 M C2H5NH2 with 265.0 mL of 0.20 M C2H5NH3Cl Express your answer using two decimal places.
Calculate the pH of a solution that results upon mixing 40ml of 0.09M NH3 with 8ml...
Calculate the pH of a solution that results upon mixing 40ml of 0.09M NH3 with 8ml of 0.18M HNO3
Calculate the pH of the solution that results from each of the following mixtures. 1)140.0 mL...
Calculate the pH of the solution that results from each of the following mixtures. 1)140.0 mL of 0.24 M HF with 230.0 mL of 0.31 M NaF 2)185.0 mL of 0.12 M C2H5NH2 with 265.0 mL of 0.21 M C2H5NH3Cl
1. Calculate the pH of a solution that results from mixing 15 mL of 0.13 M...
1. Calculate the pH of a solution that results from mixing 15 mL of 0.13 M HBrO(aq) with 11 mL of 0.1 M NaBrO(aq). The Ka value for HBrO is 2 x 10-9. 2. What is the buffer component ratio, (BrO-)/(HBrO) of a bromate buffer that has a pH of 7.91. Ka of HBrO is 2.3 x 10-9.
Calculate the pH of the solution that results from mixing 35.0 mL of 0.18 M formic...
Calculate the pH of the solution that results from mixing 35.0 mL of 0.18 M formic acid and 70.0 mL of 0.090 M sodium hydroxide. ( value for formic acid is 1.8*10-4.)
Calculate the pH of a solution made by mixing 100.0 mL of 0.050 M NH3 with...
Calculate the pH of a solution made by mixing 100.0 mL of 0.050 M NH3 with 100.0 mL of 0.100 M HCl. (K b for NH 3 = 1.8 x 10 –5 )
Calculate the pH of a solution that is 0.10M in each of the following: CH3COOH (Ka...
Calculate the pH of a solution that is 0.10M in each of the following: CH3COOH (Ka = 1.8×10-5), NaCl, HCl, and HCOOH (Ka = 1.8×10-4).
Calculate the pH at the equivalence point of a solution when 25.00mL of a 0.10M monoprotic...
Calculate the pH at the equivalence point of a solution when 25.00mL of a 0.10M monoprotic weak acid, HA, is titrated with 25.00mL of 0.10M of NaOH. The weak acid's Ka value is 1.4*10^-5.
calculate the pH of a solution that results from the addition of the following amounts of...
calculate the pH of a solution that results from the addition of the following amounts of 0.1 M NaOH to 10mL of a 0.1 HCl solution. Find the pHs for the following volumes of 0.1M NaOH added: 1, 2, 4,8, 9.8, 10.1, 10.2, 10.4, 11, 14, 18, 20 (all in mL)
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT