Chemistry, 24.12.2019 03:31 jaleewoodyard1

An analytical chemist is titrating 179.5ml of 0.8600ma solution of cyanic acid(hcno) with0.6500m a solution of naoh. the pka of cyanic acid is 3.46 . calculate the ph of the acid solution after the chemist has added 274.6ml of naoh the solution to it.

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Answer from: bloodyflower2003

12.73

Explanation:

Write the balanced equation first:

$HCNO (aq) + NaOH (aq)\rightarrow NaCNO (aq) + H_2O (l)$

Find moles of the acid (subscript of 1 for simplicity):

$n_1 = 0.8600 M\cdot 0.1795 L = 0.15437 mol$

The number of moles of NaOH at the equivalence point will be the same, as we have a 1 : 1 stoichiometry (use subscript of 2 for NaOH):

n_2 = n_1 = 0.15437 mol

Find the equivalence volume of NaOH:

$V_{eq} = \frac{n_2}{c_2} = \frac{0.15437 mol}{0.6500 M} = 237.5 mL$

Since we're looking at an instant where we actually add a total of 274.6 mL, this is much greater than the equivalence volume, so we'll have a huge excess of NaOH and we can ignore any ionization of the acid, as it'll be negligible compared to the hydroxide concentration.

Find the excess volume of NaOH added past the equivalence point:

V_2 = 274.6 mL - 237.5 mL = 37.1 mL

Find the molarity of excess NaOH using the dilution formula (we're dividing by a total volume of this solution):

$c_2 = \frac{0.6500 M\cdot 37.1 mL}{179.5 mL + 274.6 mL} = 0.053105 M$

Find the pOH value:

pOH = -log[NaOH] = -log(0.053105) = 1.27

Then:

pH = pK_w - pOH = 14.00 - 1.27 = 12.73

Answer from: Quest

periodic trends is an atom that gets larger as the number of electronic shells increase, the radius of atoms increases as you go down a certain group in the periodic table of elements. in general, the size of an atom will decrease as you move from left to right of a certain period.

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.