calculate the ph of one solution containing 0.1 M formic acid and 0.1 M sodium formate before and after the addition of 1mL of 5 M Naoh. how much could the Ph change if the NaOh were added to 1 L of pure water

Pka of formic acid (HCOOH)= 3.75

pH= PKa +log [ Sodium formate/ formic acid] = 3.75

NaOH reacts with HCOOH as HCOOH (aq) + NaOH(aq) --> NaCHO2 (aq) and H2O (l)

Moles of HCOOH= 0.1*1= 0.1 moles

Moles of NaOH= 5*1/1000= 0.005 moles

HCOOH is in excess and the excess is = 0.1 - 0.005 = 0.095

Moles of sodium formate = 0.005 + 0.1 = 0.105 moles of HCOOH= 0.095

volume after mixing = 1 + 5 /1 000=1.005

Concentrations : HCOOH= 0.095/1.005 sodium formate= 0.105/1.005

pH= 3.75+log (0.105/0.095)=3.85

When NaOH is added, molarity is , 1*5= 1000*M

M= 5/1000=0.005

The pH remains the same. Only the molarity of NaOH changes.

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.