Ineed some with titrations and need to have someone check my answers.

1. if it takes 67 ml of 0.15 m naoh to neutralize 134 ml of an hcl solution, what is the concentration of the hcl?

the concentration is 0.075m of hcl

2. if it takes 27.4 ml of 0.050 m h2so4 to neutralize 357 ml of koh solution, what is the concentration of the naoh solution?

it takes 0.05m of koh. now i'm not sure if the naoh is there as a typo but is it possible i'm supposed to figure out the concentration of naoh from h2so4 and koh? i don't know too much about chemistry but when i balanced it out i didn't get that so i'm not sure.

and here's the last one:

3. if it takes 55 ml of 0.5 m naoh solution to completely neutralize 130 ml of sulfuric acid solution (h2so4), what is the concentration of the h2so4 solution?

the concentration of h2so4 is 0.5m

are my answers correct?

1. This is an acid base neutralisation reaction ,

the balanced equation for the reaction is as follows

NaOH + HCl ---> NaCl + H₂O

molar ratio of NaOH to HCl is 1:1

number of moles are calculated by concentration x volume

therefore at neutralisation number of moles of HCl reacted is equal to number of NaOH moles

number of moles of NaOH reacted = 0.15 mol/dm³ x 67 x 10⁻³ dm³ = 10.05 x10⁻³ mol

the number of moles of HCl in 134 x 10⁻³ dm³ is 10.05 x 10⁻³ mol

to find the concentration of HCl , we need to find the number of moles in 1 dm³

in 134 x 10⁻³ dm³ there are - 10.05 x 10⁻³ mol

therefore in 1 dm³ there are- 10.05 x 10⁻³ mol / 134 x 10⁻³ dm³ = 0.075 mol/dm³  

2. the acid base neutralisation reaction between KOH and H₂SO₄ is

2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O

molar ratio of KOH to H₂SO₄ is 2:1

1 mol of H₂SO₄ reacts with 2 mol of KOH

the number of H₂SO₄ moles reacted - 0.050 mol/dm³ x 27.4 x 10⁻³ dm³ = 1.37 x 10⁻³ mol

number of KOH moles reacted is twice the number of H₂SO₄ moles reacted according to molar ratio

therefore number of HCl moles are - 1.37 x 10⁻³ x 2 = 2.74 x 10⁻³ mol

there are 2.74 x 10⁻³ mol of KOH in 357 x 10⁻³ dm³

therefore in 1 dm³ of KOH there are - 2.74 x 10⁻³ mol / 357 x 10⁻³ dm³

the concentration of KOH is - 0.00768 mol/dm³

even though its NaOH or KOH the concentration would be the same.  

3. the balanced equation is

2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

molar ratio of NaOH to H₂SO₄ is 2:1

2 mol of NaOH reacts with 1 mol of H₂SO₄

number of NaOH moles reacted - 0.5 mol/dm³ x 55 x 10⁻³ dm³ = 25.5 x 10⁻³ mol

number of H₂SO₄ moles reacted is - half the number of NaOH moles reacted

number of H₂SO₄ moles reacted - 25.5 x 10⁻³ mol / 2 = 12.75 x 10⁻³ mol

there are 12.75 x 10⁻³ mol in 130 x 10⁻³ dm³

then in 1 dm³ there are - 12.75 x 10⁻³ mol / 130 x 10⁻³ dm³ = 0.098 mol/dm³

concentration is 0.098 M

the answer for sodium is 0.186

the answer for rubidium is 0.251

b is the correct answer

mark me brainliest and hope this