the pka value for hso3- is 7.19. what mole ratio o
the pka value for hso3- is 7.19. what mole ratio of na2so3 to nahso3 is needed to prepare a buffer with a ph of 7.39? [so3^2-] / [hso3-]
since the solution is a buffer, we can use the henderson-hasselbalch equation to calculate the ratio of [so32-]/[hso3-] to have a ph of 7.39:ph=pka+logconjugate baseweak acidin the solution, identify the weak acid and the conjugate base. recall from the bronsted-lowry definition that an acid is a proton (h+) donor and the base is a proton (h+) acceptor.• nahso3 is a salt that will give hso3- when it dissociatesnahso3 → na+ + hso3-• na2so3 is a salt that will give so32- when it dissociatesna2so3 → 2 na+ + so32-•hso3- is a weak acid.          - concentration of hso3- = concentration of nahso3• when hso3- loses h+ to give a conjugate base, so32- is formedhso32- - h+ → so32-- concentration of so32- = concentration of na2so3      hso3-    +   h2o     →        so32-        +       h3o+(weak acid)    (base)      (conjugate base)  (conjugate acid)we can rewrite the henderson-hasselbalch equation as:ph=pka+logso32-hso3-calculate [so32-]/[hso3-] ratio:• ph = 7.39• pka = 7.19ph=pka+logso32-hso3-7.39=7.19+logso32-hso3-logso32-hso3-=7.39-7.19logso32-hso3-log=0.2loglog has a base of 10 and when you divide a number by log you will take 10 to the power of the numberso32-hso3-=100.2[so32-]/[hso3-] = 1.58the ratio [so32-]/[hso3-] is 1.58
a) ans. concentration = no. of moles of solute * 1000/ vol. in m;

no. of moles of solute = concentration * vol / 1000 = 1.4 * 270 / 1000 = 0.378 = 0.378 mol.



b)

molality = 1.65

molality = moles of solute / mass of solvent in kg

1.65 = moles of solute / 5.5 x 10^-5 kg

moles of solute = 9.075 x 10^-5

c)

mass of sucrose = 75 x 1.4 / 100 = 1.05g

moles = 1.05 / 342 = 3.07 x 10^-3 AgNO3(aq) + NaCl(s) = AgCl(s) + NaNO3  
n(AgCl) = 0.765 g / M(AgCl)  
M(AgCl) = 108 + 35.5 = 143.5 g/mol  
n(AgCl) = 0.765 g /143.5 g/mol = 0.00533 mol  
n(AgCl) = chloride ions = 0.00533 mol  
C(chloride ions) = n/V  
C(chloride ions) = 0.00533 mol/1l = 0.00533 mol/ L (a)  
m(NaCl)=n(NaCl)*M(NaCl)  
n(NaCl) = 0.00533 mol  
M(NaCl) = 23 + 35.5 = 58.5 g/mol  
m(NaCl) = 0.00533 mol * 58.5 g/mol = 0.3115 g (b)  
  
**Answer:**  
**a) 0.00533 mol/ L**  
**b) 0.3115 g** NaBr   

  

SO3   

  

CaCl2   

  

Mg3N2 because of the small size of boron atom and presence of only six electrons in its valence shell, b(oh)3 accepts a pair of electrons from oh- ion of h2o, releasing a proton.the reaction mechanism is: a) 3ag + 4hn03 > 3agno3 + no + 2h2o

b) mno2 + 4hcl > 2h2o + mncl2 + cl2 V(NaOH) = 2×9×0.122/0.141 = 15.57 mL use the ka equation: 
take the acid as ha
this will dissociate : 
ha ↔ h+ + a- 
ka = [h+] [a-] /[ha] 
because dissociation is very small we can use 0.1m for [ha] 
[h+] = 2/50*0.1 = 0.0004m 
[a-] = 0.0004m 
 ka 0.0004 * 0.0004 / 0.1 
ka = 1.6*10^-5 * 0.1 
ka = 1.6*10^-4 have added psu to above post. (i added specs because i knew someone would ask. however i think it's a software issue.)

unlikely that it's the psu.

the current psu is a hx750i, and the previous psu was a hx650i. swapped over a year ago at least. this is a recent issue.


good, a lot of times this is the psu. but with an excellent unit, that's far less likely.

were you monitoring temperatures at the time? was there something specific you were doing similar in both times? really intermittent things can be hard to nail down. First, you want to convert grams of Iron Hydroxide into moles like this: (55.0g/89.85 g mol^-1) which is 0.6121313 moles.   

  

For each mole of Iron Hydroxide there are 2 moles of oxygen, so you multiply 0.6121313 by 2 to get 1.2242626.   

  

Then, you can use Avogadro's number to convert to number of atoms by multiplying 1.2242626 by 6.0221415x10^23 to get your final answer.   

  

Good luck. both products can affect your kidneys. use either up to 3,000mg of tylenol or 1500 mg/day of naproxyn (naprosyn) sodium ( aleve). 7-10 days is a good beginning duration and have your pcp do blood and urine testing to check your kidney function. see more...